AN 


ELEMENTARY    TREATISE 


ON 


ANALYTIC  MECHANICS 


WITH    NUMEROUS     EXAMPLES. 


BY 

EDWARD   A.   BOWSER,   LL.D., 

PROFESSOR  OP   MATHEMATICS   AND    ENGINEERING   IN  RUTGERS  COLLEGE 


TWENTY-SECOND  Bp^TKW:. 


NEW   YORK: 

D.    VAN   NOSTRAND    COMPANY, 

25  Park  Place, 

1912 


i0< 
J- 


COPYRIGHT,  188k,  BY E.  A.  BOWSES. 


PREFACE. 


■»♦» 


TT^HE  present  work  on  Analytic  Mechanics  or  Dynamics  is  designed 
as  a  text-book  for  the  students  of  Scientific  Schools  and  Col- 
leges, who  have  received  training  in  the  elements  of  Analytic  Geome- 
try and  the  Calculus. 

Dynamics  is  here  used  in  its  true  sense  as  the  science  of  force. 
The  tendency  among  the  best  and  most  logical  writers  of  the  present 
day  appears  to  be  to  use  this  term  for  the  science  of  Analytic  Me- 
chanics, while  the  branch  formerly  called  Dynamics  is  now  termed 
Kinetics. 

The  treatise  is  intended  especially  for  beginners  in  this  branch  of 
science.  It  involves  the  use  of  Analytic  Geometry  and  the  Calculus. 
The  analytic  method  has  been  chiefly  adhered  to,  as  being  better 
adapted  to  the  treatment  of  the  subject,  more  general  in  its  applica- 
tion and  more  fruitful  in  results  than  the  geometric  method;  and  yet 
where  a  geometric  proof  seemed  preferable  it  has  been  introduced. 

The  aim  has  been  to  make  every  principle  clear  and  intelligible, 
to  develop  the  different  theories  with  simplicity,  and  to  explain  fully 
the  meaning  and  use  of  the  various  analytic  expressions  in  which  the 
principles  are  embodied. 

The  book  consists  of  three  parts.  Part  I,  with  the  exception  of  a 
preliminary  chapter  devoted  to  definitions  and  fundamental  princi- 
ples, is  entirely  given  to  Statics. 

Part  II  is  occupied  with  Kinematics,  and  the  principles  of  this 
important  branch  of  mathematics  are  so  treated  that  the  student  may 
enter  upon  the  study  of  Kinetics  with  clear  notions  of  motion,  veloc- 
ity and  acceleration.  Part  III  treats  of  the  Kinetics  of  a  particle  and 
of  rigid  bodies.  O  fj  *y  Q  A  J 


IV  PRE  FA  CE. 

In  this  arrangement  of  the  work,  with  the  exception  of  Kine- 
matics, I  have  followed  the  plan  usually  adopted,  and  made  the 
subject  of  Statics  precede  that  of  Kinetics. 

For  the  attainment  of  that  grasp  of  principles  which  it  is  the 
special  aim  of  the  book  to  impart,  numerous  examples  are  given  at 
the  ends  of  the  chapters.  The  greater  part  of  them  will  present  no 
serious  difficulty  to  the  student,  while  a  few  may  tax  his  best 
efforts. 

In  preparing  this  book  I  have  availed  myself  of  the  writings  of 
many  of  the  best  authors.  The  chief  sources  from  which  I  have 
derived  assistance  are  the  treatises  of  Price,  Minchin,  Todhunter, 
Pratt,  Routh,  Thomson  and  Tait,  Tait  and  Steele,  Weisbach,  Ventu- 
roli,  Wilson,  Browne,  Gregory,  Rankine,  Boucharlat,  Pirie,  Lagrange 
and  La  Place,  while  many  valuable  hints  as  well  as  examples  have  been 
obtained  from  the  works  of  Smith,  Wood,  Bartlett,  Young,  Moseley, 
Tate,  Magnus,  Goodeve,  Parkinson,  Olmsted,  Garnett,  Renwick,  Bot- 
tomley,  Morin,  Twisden,  Whewell,  Galbraith,  Ball,  Dana,  Byrne,  the 
Encyclopaedia  Britannica,  and  the  Mathematical  Visitor. 

I  have  again  to  thank  my  old  pupil,  Mr.  R.  W.  Prentiss,  of  the 
Nautical  Almanac  Office,  and  formerly  Fellow  in  Mathematics  at  the 
Johns  Hopkins  University,  for  reading  the  MS.  and  for  valuable  sug- 
gestions. Several  others  also  of  my  friends  have  kindly  assisted  me 
by  correcting  proof-sheets  and  verifying  copy  and  formulae. 

E.  A.  B. 
Rutgers  College,  ) 

New  Brunswick,  N.  J.,  June,  1884. » 


TABLE   OF   CONTENTS 


PART    I 


CHAPTER    I. 

FIRST    PRINCIPLES. 

ABT.  PAOB 

1.  Definitions — Statics,  Kinetics  and  Kinematics 1 

2.  Matter 2 

3.  Inertia ..   2 

4.  Body,  Space  and  Time 2 

5.  Rest  and  Motion 2 

6.  Velocity 3 

7.  Formulae  for  Velocity 4 

8.  Acceleration. ...           4 

9.  Measure  of  Acceleration 5 

10.  Geometric  Representation  of  Velocity  and  Acceleration 6 

11.  Mass 6 

12.  Momentum 7 

13.  Change  of  Momentum 7 

14.  Force 8 

15.  Static  Measure  of  Force 8 

16.  Action  and  Reaction 9 

17.  Method  of  Comparing  Forces 9 

18.  Representation  of  Forces 10 

19.  Measure  of  Accelerating  Forces 10 

20.  Kinetic  Measure  of  Force 11 

21 .  Absolute  or  Kinetic  Unit  of  Force 13 

22.  Three  Ways  of  Measuring  Force 14 

23.  Meaning  of  g  in  Dynamics 15 


VI  CONTENTS. 

ART.  PAGE 

24.  Gravitation  Units  of  Force  and  Mass 1G 

25.  Gravitation  and  Absolute  Measure 17 

Examples 13 


STATICS     (REST). 


CHAPTER    II. 

THE      COMPOSITION      AND      RESOLUTION      OF     CONCURRING 
FORCES — CONDITIONS    OF    EQUILIBRIUM. 

26.  Problem  of  Statics 21 

27.  Concurring  and  Conspiring  Forces 21 

28.  Composition  of  Conspiring  Forces 22 

29.  Composition  of  Velocities 28 

30.  Composition  of  Forces 24 

31.  Triangle  of  Forces 25 

32.  Three  Concurring  Forces  in  Equilibrium. 26 

33.  The  Polygon  of  Forces 27 

34.  Parallelopiped  of  Forces 28 

35.  Resolution  of  Forces 30 

36.  Magnitude  and  Direction  of  Resultant 31 

37.  Conditions  of  Equilibrium 32 

38.  Resultant  of  Concurring  Forces  in  Space 34 

39.  Equilibrium  of  Concurring  Forces  in  Space 35 

40.  Tension  of  a  String 35 

41.  Equilibrium  of  Concurring  Forces  on  a  Smooth  Plane 39 

42.  Equilibrium  of  Concurring  Forces  on  a  Smooth  Surface 41 

Examples 45 


CHAPTER    III. 

COMPOSITION    AND    RESOLUTION    OF  FORCES    ACTING   ON  A 
RIGID  BODY. 

43.  ARigidBody 57 

44.  Transmissibility  of  Force 57 

45.  Resultant  of  Two  Parallel  Forces 58 

46.  Moment  of  a  Force 60 


CONTENTS.  Til 

*RT.  PAQB 

47.  Signs  of  Moments 01 

48.  Geometric  Representation  of  a  Moment 01 

49.  Two  Equal  and  Opposite  Parallel  Forces 01 

50.  Moment  of  a  Couple 02 

51.  Effect  of  a  Couple  on  a  Rigid  Body 03 

52.  Effect  of  Transferring  Couple  to  Parallel  Plane  not  altered. . .  04 
53    A  Couple  replaced  by  another  Couple G4 

54.  A  Force  and  a  Couple.  05 

55.  Resultant  of  any  number  of  Couples 00 

56.  Resultant  of  Two  Couples 67 

57.  Varignon's  Theorem  of  Moments 09 

58.  Varignon's  Theorem  for  Parallel  Forces 71 

59.  Centre  of  Parallel  Forces 71 

60.  Equilibrium  of  a  Rigid  Body  under  Parallel  Forces 74 

61.  Equilibrium  of  a  Rigid  Body  under  Forces  in  any  Direction. .  75 

62.  Equilibrium  under  Three  Forces 77 

63.  Centre  of  Parallel  Forces  in  Different  Planes 85 

64.  Equilibrium  of  Parallel  Forces  in  Space 86 

65.  Equilibrium  of  Forces  acting  in  any  Direction  in  Space 88 

Examples 90 


CHAPTER    IV. 

CENTRE   OF   GRAVITY    (CENTRE   OF   MASS). 

66.  Centre  of  Gravity 100 

67.  Planes  of  Symmetry — Axes  of  Symmetry 101 

68.  Body  Suspended  from  a  Point 101 

69.  Body  Supported  on  a  Surface 102 

70.  Different  Kinds  of  Equilibrium 102 

71.  Centre  of  Gravity  of  Two  Masses 103 

72.  Centre  of  Gravity  of  Part  of  a  Body 103 

73.  Centre  of  Gravity  of  a  Triangle 103 

74.  Centre  of  Gravity  of  a  Triangular  Pyramid 105 

75.  Centre  of  Gravity  of  a  Cone 106 

76.  Centre  of  Gravity  of  Frustum  of  Pyramid 107 

77.  Investigations  involving  Integration 109 

78.  Centre  of  Gravity  of  the  Arc  of  a  Curve 110 

79.  Centre  of  Gravity  of  a  Plane  Area 115 

80.  Polar  Elements  of  a  Plane  Area 118 


Vlll  CONTENTS. 

AKT  PAGE 

81.  Double  Integration — Polar  Formulae 120 

82.  Double  Integration — Rectangular  Formulae 122 

83.  Centre  of  Gravity  of  a  Surface  of  Revolution 123 

84.  Centre  of  Gravity  of  any  Curved  Surface 126 

85.  Centre  of  Gravity  of  a  Solid  of  Revolution 127 

86.  Polar  Formulas 130 

87.  Centre  of  Gravity  of  any  Solid 131 

88.  Polar  Elements  of  Mass 133 

89.  Special  Methods 136 

90.  Theorems  of  Pappus 138 

Examples 140 


CHAPTER    V. 

FRICTION. 

91.  Friction 149 

92.  Laws  of  Friction 150 

93.  Magnitudes  of  Coefficients  of  Friction 152 

94.  Angle  of  Friction 152 

95.  Reaction  of  a  Rough  Curve  or  Surface 153 

96.  Friction  on  an  Inclined  Plane 154 

97.  Friction  on  a  Double  Inclined  Plane 156 

98.  Friction  on  Two  Inclined  Planes 159 

99.  Friction  of  a  Trunnion 159 

100.  Friction  of  a  Pivot 160 

Examples 162 


CHAPTER    VI. 

THE   PRINCIPLE   OF  VIRTUAL  VELOCITIES. 

101.  Virtual  Velocity f  <56 

102.  Principle  of  Virtual  Velocities 167 

103.  Nature  of  the  Displacement 169 

104.  Equation  of  Virtual  Moments 169 

105.  System  of  Particles  Rigidly  Connected 170 

Examples 172 


CONTENTS.  ix 


CHAPTER    VII. 

MACHINES. 

*KT.  PAOB 

106.  Functions  of  a  Machine 177 

107.  Mechanical  Advantage 178 

108.  Simple  Machines 180 

109.  The  Lever 181 

110.  Equilibrium  of  the  Lever 181 

111.  The  Common  Balance 184 

112.  Chief  Requisites  of  a  Good  Balance 186 

113.  The  Steelyard 188 

114.  To  Graduate  the  Common  Steelyard 188 

115.  The  Wheel  and  Axle 190 

116.  Equilibrium  of -the  Wheel  and  Axle 190 

117.  Toothed  Wheels 192 

118.  Relation  of  Power  and  Weight  in  Toothed  Wheels 193 

119.  Relation  of  Power  to  Weight  in  a  Train  of  Wheels 194 

120.  The  Inclined  Plane 196 

121.  The  Pulley 197 

122.  The  Simple  Movable  Pulley 198 

123.  First  System  of  Pulleys 198 

124.  Second  System  of  Pulleys 200 

125.  Third  System  of  Pulleys 201 

126.  The  Wedge 202 

127.  Mechanical  Advantage  of  the  Wedge 202 

128.  The  Screw 204 

129.  Relation  between  Power  and  Weight  in  the  Screw 204 

129a.  Prony's  Differential  Screw .206 

Examples 207 


CHAPTEE    VIII. 

THE   FUNICULAR  POLYGON — THE   CATENARY — ATTRACTION 

130.  Equilibrium  of  the  Funicular  Polygon 216 

131.  To  Construct  the  Funicular  Polygon 218 

132.  Cord  Supporting  a  Load  Uniformly  Distributed 219 

133.  The  Common  Catenary — Its  Equation 221 

133a.  Attraction  of  a  Spherical  Sliell 226 

Examples 229 


CONTENTS, 


PART   II. 
KINEMATICS     (MOTION). 


CHAPTER    I. 

RECTILINEAR     MOTION". 

hi??.  PAGR 

134.  Definitions— Velocity 231 

135.  Acceleration 233 

136.  Relation  between  Space  and  Time  when  Acceleration  =  0.. .  233 

137.  Relation  when  the  Acceleration  is  Constant 234 

138.  Relation  when  Acceleration  varies  as  the  Time 235 

139.  Relation  when  Acceleration  Varies  as  the  Distance 235 

140.  Equations  of  Motion  for  Falling  Bodies 237 

141.  Particle  Projected  Vertically  Upwards 239 

142.  Compositions  of  Velocities 242 

143.  Resolution  of  Velocities 243 

144.  Motion  on  an  Inclined  Plane 245 

145.  Times  of  Descent  down  Chords  of  a  Circle 247 — 

146.  The  Straight  Line  of  Quickest  Descent 248 

Examples 249 


CHAPTER    II. 

CURVILINEAR    MOTION. 

147.  Remarks  on  Curvilinear  Motion 258 

148.  Composition  of  Uniform  Velocity  and  Acceleration 258 

149.  Composition  and  Resolution  of  Acceleration 259 

Examples ; * .  261 

150.  Motion  of  Projectiles  in  Vacuo 266 

151.  The  Path  of  a  Particle  in  Vacuo  is  a  Parabola 266 

152.  The  Parameter — Range — Greatest  Height — Height  of  Direc- 

trix   267 

153.  Velocity  of  a  Particle  at  any  point  of  its  Path 269 

154.  Time  of  Flight  along  a  Horizontal  Plane 269 

155.  Point  at  which  a  Projectile  will  Strike  an  Inclined  Plane. . .  270 


CONTENTS.  Xi 

AW.  PAGE 

156.  Projection  for  Greatest  Range  on  a  Given  Plane 270 

157.  The  Elevation  that  the  Particle  may  pass  a  (liven  Point. . .  271 

158.  Second  Method  of  Finding  Equation  of  Trajectory 272 

159.  Velocity  of  Discharge  of  Balls  and  Shells 274 

160.  Angular  Velocity  and  Angular  Acceleration 275 

161.  Accelerations  Along  and  Perpendicular  to  Radius  Vector. . . .  278 

162.  Accelerations  Along  and  Perpendicular  to  Tangent 279 

163.  When  Acceleration  Perpendicular  to  Radius  Vector  is  zero. .  281 

164.  When  Angular  Velocity  is  Constant 282 

Examples 284 


PART    III. 
KINETICS     (MOTION    AND    FORCE) 


CHAPTER    I. 

LAWS     OF     MOTION — MOTION     UNDER     THE     ACTION     OF     A 
VARIABLE   FORCE — MOTION   IN   A   RESISTING   MEDIUM. 

165.  Definitions 289 

166.  Newton's  Laws  of  Motion 289 

167.  Remarks  on  Law  1 290 

168.  Remarks  on  Law  II.. 291 

169.  Remarks  on  Law  III 294 

170.  Two  Laws  of  Motion  in  the  French  Treatises 295 

171.  Motion  of  Particle  under  an  Attractive  Force 295 

172.  Motion  under  the  action  of  a  Variable  Repulsive  Force 298 

173.  Motion  under  the  action  of  an  Attractive  Force 299 

174.  Velocity  acquired  in  Falling  through  a  Great  Height 300 

175.  Motion  in  a  Resisting  Medium.   302 

176.  Motion  in  the  Air  against  the  Action  of  Gravity 304 

177.  Motion  of  a  Projectile  in  a  Resisting  Medium 307 

178.  Motion  against  the  Resistance  of  the  Atmosphere 308 

179.  Motion  in  the  Atmosphere  under  a  small  Angle  of  Elevation  312 
Examples 313 


HI  CONTENTS, 


CHAPTER    II. 

CENTRAL  FORCES. 

AST.  PAGB 

180.  Definitions 321 

181.  A  Particle  under  the  Action  of  a  Central  Attraction 321 

182.  The  Sectorial  Area  Swept  over  by  the  Radius  Vector 325 

183.  Velocity  of  Particle  at  any  Point  of  its  Orbit 325 

184.  Orbit  when  Attraction  as  the  Inverse  Square  of  Distance. . .  329 

185.  Suppose  the  Orbit  to  be  an  Ellipse 333 

186.  Kepler's  Laws 335 

187.  Nature  of  the  Force  which  acts  upon  the  Planetary  System.  335 
Examples 338 


CHAPTEE    III. 

CONSTRAINED    MOTION 

188.  Definitions 345 

189.  Kinetic  Energy  or  Vis  Viva — Work 345 

190.  To  Find  the  Reaction  of  the  Constraining  Curve 348 

191.  Point  where  Particle  will  leave  Constraining  Carve 349 

192.  Constrained  Motion  Under  Action  of  Gravity 350 

193.  Motion  on  a  Circular  Arc  in  a  Vertical  Plane 350 

194.  The  Simple  Pendulum .1352^ 

195.  Relation  of  Time,  Length,  and  Force  of  Gravity  .     353 

196.  Height  of  Mountain  Determined  with  Pendulum 354 

197.  Depth  of  Mine  Determined  with  Pendulum 355 

198.  Centripetal  and  Centrifugal  Forces 356 

199.  The  Centrifugal  Force  at  the  Equator 358 

200.  Centrifugal  Force  at  Different  Latitudes 359 

201.  The  Conical  Pendulum— The  Governor. 361 

Examples 362 

CHAPTER    IV. 

IMPACT. 

202.  An  Impulsive  Force 370 

203.  Impact  or  Collision 371 

204.  Direct  and  Central  Impact 372 

205.  Elasticity  of  Bodies— Coeffi  jicnt  of  Restitution 373 


CONTENTS.  XI  li 

ART.  FAOE 

206.  Direct  Impact  of  Inelastic  Bodies 374 

207.  Direct  Impact  of  Elastic  Bodies 375 

208.  Loss  of  Kinetic  Energy  in  Impact  of  Bodies 378 

209.  Oblique  Impact  of  Bodies 380 

210.  Oblique  Impact  of  Two  Smooth  Spheres 382 

Examples 383 

CHAPTER    V. 

WORK   AND   ENERGY. 

211.  Definition  and  Measure  of  Work 389 

212.  General  Case  of  Work  done  by  a  Force 390 

213.  Work  on  an  Inclined  Plane. 391 

Examples 393 

214.  Horse  Power 395 

215.  Work  of  Raising  a  System  of  Weights 396 

Examples 397 

216.  Modulus  of  a  Machine 400 

Examples 401 

217.  Kinetic  and  Potential  Energy— Stored  Work 404 

Examples 406 

218.  Kinetic  Energy  of  a  Rigid  Body  Revolving  round  an  Axis. . .  408 

219.  Force  of  a  Blow 411 

220.  Work  of  a  Water  Fall 412 

221.  The  Duty  of  an  Engine 414 

222.  Work  of  a  Variable  Force 415 

223.  Simpson's  Rule 415 

Examples 417 

CHAPTER    VI. 

MOMENT  OF   INERTIA. 


224.  Moment  of  Inertia 429 

225.  Moments  of  Inertia  relative  to  Parallel  Axes  or  Planes 432 

226.  Radius  of  Gyration 434 

227.  Polar  Moment  of  Inertia 436 

228.  Moment  of  Inertia  of  a  Solid  of  Revolution _437_ 

229.  Moment  of  Inertia  about  Axis  Perpendicular  to  Geometric 

Axis 438 

230.  Moment  of  Inertia  of  Various  Solid  Bodies 440 

231.  Moment  of  Inertia  of  a  Lamina  with  respect  to  any  Axis 44* 


29  k  V 

:34        ) 
36    j 


XIV  CONTENTS, 

AKT.  pAGK 

232.  Principal  Axes  of  a  Body 442 

233.  Products  of  Inertia 446 

Examples 447 

CHAPTER    VII. 

ROTATORY   MOTION". 

234.  Impressed  and  Effective  Forces 451 

235.  D'Alembert's  Principle 452 

236.  Rotation  of  a  Rigid  Body  about  a  Fixed  Axis 454 

237.  The  Compound  Pendulum 457 

238.  Length  of  Second's  Pendulum  Determined  Experimentally. .  462 

239.  Motion  of  a  Body  when  Unconstrained 464 

240.  Centre  of  Percussion — Axis  of  Spontaneous  Rotation 464 

241.  Principal  Radius  of  Gyration  Determined  Practically 467 

242.  The  Ballistic  Pendulum 468 

243.  Motion  of  a  Body  about  a  Horizontal  Axle  through  its  Centre  470 

244.  Motion  of  a  Wheel  and  Axle 471 

245.  Motion  of  a  Rigid  Body  about  a  Vertical  Axis 472 

246.  Body  Rolling  down  an  Inclined  Plane 473 

247.  Falling  Body  under  an  Impulse  not  through  its  Centre 475 

Examples 477 

CHAPTER    VIII. 

MOTION   OF  A   SYSTEM   OF   RIGID   BODIES   IN"   SPACE. 

248.  Equations  of  Motion  obtained  by  D'Alembert's  Principle  . . .  481 

249.  Independence  of  the  Motions  of  Translation  and  Rotation. . .  482 

250.  Principle  of  the  Conservation  of  the  Centre  of  Gravity 485 

251.  Principle  of  the  Conservation  of  Areas 486 

252.  Conservation  of  Vis  Viva  or  Energy 488 

253.  Composition  of  Rotations 493 

254.  Motion  of  a  Rigid  Body  referred  to  Fixed  Axes 494 

255.  Axis  of  Instantaneous  Rotation 495 

256.  Angular  Velocity  about  Axis  of  Instantaneous  Rotation 496 

257.  Euler's  Equations 497 

258.  Motion  about  a  Principal  Axis  through  Centre  of  Gravity. . .  499 

259.  Velocity  about  a  Principal  Axis  when  Accelerating  Forces 

=  0 501 

S60.  The  Integral  of  Euler's  Equations 502 

Examples 505 


ANALYTIC    MECHANICS, 


PART    I. 


CHAPTER     I. 

FIRST    PRINCIPLES 


1.  Definitions. — Analytic   Mechanics  or  Dynamics  is 

the  science  which  treats  of  the  equilibrium  and  motion 
of  bodies  under  the  action  of  force.  It  is  accordingly 
divided  into  two  parts,  Statics  and  Kinetics. 

Statics  treats  of  the  equilibrium  of  bodies,  and  the  condi- 
tions governing  the  forces  which  produce  it. 

Kinetics  treats  of  the  motion  of  bodies,  and  the  laws  of 
the  forces  which  produce  it. 

The  consideration  that  the  properties  of  motion,  velocity, 
and  displacement  may  be  treated  apart  from  the  particular 
forces  producing  them  and  independently  of  the  bodies  sub- 
ject to  them,  has  given  rise  to  an  auxiliary  branch  of  Dyna- 
mics called  Kinematics.* 

Although  Kinematics  may  not  be  regarded  as  properly 
included  under  Dynamics,  yet  this  branch  of  science  is  so 
important  and  useful,  and  its  appl/cation  to  Dynamics  so 
immediate,  that  a  portion  of  this  work  is  devoted  to  its 
treatment. 

*  This  name  was  given  by  Ampere. 


2  MATTER,   INERTIA,    BODY,   MOTION,    ETC. 

Kinematics  1*  the  science  of  pure  motion,  without  refer- 
ence to  matter  or  force.  It  treats  of  the  properties  of 
motion  without  regard  to  what  is  moving  or  how  it  is 
moved.  It  is  an  extension  of  pure  geometry  by  introduc- 
ing the  idea  of  time,  and  the  consequent  idea  of  velocity. 

2.  Matter.— Matter  is  that  which  can  be  perceived  by 
the  senses,  and  which  can  transmit,  and  be  acted  upon  by 
force.     It  has  extension,  resistance,  and  impenetrability. 

A  definition  of  matter  which  would  satisfy  the  metaphysician  is 
not  required  for  this  work.  It  is  sufficient  for  us  to  conceive  of  it  as 
capable  of  receiving  and  transmitting  force  ;  because  it  is  in  this 
aspect  only  that  it  is  of  importance  in  the  present  treatise. 

3.  Inertia. — By  Inertia  is  meant  that  property  of  mat- 
ter by  which  it  remains  in  its  state  of  rest  or  uniform 
motion  in  a  right  line  unless  acted  upon  by  force.  Inertia 
expresses  the  fact  that  a  body  cannot  of  itself  change  its 
condition  of  rest  or  motion.  It  follows  that  if  a  body 
change  its  state  from  rest  to  motion  or  from  motion  to  rest, 
or  if  it  change  its  direction  from  the  natural  rectilinear 
path,  it  must  have  been  influenced  by  some  external  cause. 

4.  Body,  Space,  and  Time.—  A  Body  is  a  portion  of 
matter  limited  in  every  direction,  and  is  consequently  of  a 
determinate  form  and  volume. 

A  Rigid  Body  is  one  in  which  the  relative  positions  of 
its  particles  remain  unchanged  by  the  action  of  forces. 

A  Particle  is  a  body  indefinitely  small  in  every  direction, 
and  though  retaining  its  material  properties  may  be  treated 
as  a  geometric  point. 

Space  is  indefinite  extension.  Time  is  any  limited  por- 
tion of  duration. 

5.  Rest  and  Motion.  —A  body  is  at  rest  when  it  con- 
stantly  occupies    : .he  same    place  in  space.     A  body  is  in 


VELOCITY.  3 

motion  when  the  body  or  its  parte  occupy  successively  dif- 
ferent positions  in  space.  But  we  cannot  judge  of  the  state 
of  rest  or  motion  of  a  body  without  referring  it  to  the 
positions  of  other  bodies  ;  and  hence  rest  and  motion  musl 

be  considered  as  necessarily  via  I  ice. 

If  tliere  were  anything  which  we  knew  to  be  absolutely  fixed  in 
space,  we  might  perceive  absolute  motion  by  change  of  place  with 
reference  to  that  object.  But  as  we  know  of  no  such  thing  as  abso- 
lute rest,  it  follows  that  all  motion,  as  measured  by  us,  must  be 
relative ;  i.  e.,  must  relate  to  something  which  we  assume  to  be  fixed. 
Hence  the  same  thing  may  often  be  said  to  be  at  rest  and  in  motion 
at  the  same  time  ;  for  it  may  be  at  rest  in  regard  to  one  thing,  and  in 
motion  in  regard  to  another.  For  example,  the  objects  on  a  vessel 
may  be  at  rest  with  reference  to  each  other  and  to  the  vessel,  while  they 
are  in  motion  with  reference  to  the  neighboring  shore.  So  a  man, 
punting  his  barge  up  the  river,  by  leaning  against  a  pole  which  rests 
on  the  bottom,  and  walking  on  the  deck,  is  in  motion  relative  to  the 
barge,  and  in  motion,  but  in  a  different  manner,  relative  to  the  cur- 
Tent,  while  he  is  at  rest  relative  to  the  earth. 

Motion  is  uniform  when  the  body  passes  over  equal  spaces 
in  equal  times  ;  otherwise  it  is  variable. 

6.  Velocity. —  The  velocity  of  a  bod//  is  its  rate  of 
motion.  When  the  velocity  is  constant,  it  is  measured  by 
the  space  passed  over  in  a  unit  of  time.  When  it  is  varia- 
ble, it  is  measured,  at  any  instant,  by  the  space  over  which 
the  body  would  pass  in  a  unit  of  time,  were  it  to  move, 
during  that  unit,  with  the  same  velocity  that  it  has  at  the 
instant  considered. 

The  speed  of  a  railway  train  is,  in  general,  variable.  If  we  were  to 
say,  for  example,  that  it  was  running  at  the  rate  of  30  miles  an  hour, 
we  would  not  mean  that  it  ran  80  miles  during  the  last  hour,  nor  that 
it  would  run  30  miles'during  the  next  hour.  We  would  mean  that,  if 
it  were  to  run  for  an  hour  with  the  speed  which  it  now  has,  at  the 
instant  considered,  it  would  pass  over  exactly  o0  miles. 

In  order  to  have  a  uniform  unit  of  velocity,  it  is  custom- 
ary to  express  it  in  feet  and  seconds ;  and  when  velocities 


4  ACCELERATION. 

are  expressed  in  any  other  terms,  they  should  be  reduced  to 
their  equivalent  value  in  feet  and  seconds.  The  unit 
velocity,  therefore,  is  the  velocity  with  which  a  body 
describes  one  foot  in  one  second  ;  other  units  may  be  taken 
where  convenience  demands,  as  miles  and  hours,  etc. 

When  we  speak  of  the  space  passed  over  by  a  body,  we 
mean  the  path  or  line  which  a  point  in  the  body  or  which  a 
particle  describes. 

7.  Formulae  for  Velocity. — If  s  be  the  space  passed 
over  by  a  particle  in  t  units  of  time,  and  v  the  velocity,  it  is 
plain  that,  for  uniform  velocity,  we  shall  have 

—  }i  a) 

that  is,  we  divide  the  whole  space  passed  over  by  the  time 
of  the  motion  over  that  space. 

If  the  velocity  continually  changes,  equal  increments  are 
not  described  in  equal  times,  and  the  velocity  becomes 
a  function  of  the  time.  But  however  much  the  velocity 
changes,  it  may  be  regarded  as  constant  during  the 
infinitesimal  of  time  dt,  in  which  time  the  body  will 
describe  the  infinitesimal  of  space  ds.  Hence,  denoting  the 
velocity  at  any  instant  by  v,  we  have 

v  =  s-  .  (3) 

In  this  case  the  velocity  is  the  ratio  of  two  infinitesimals. 
These  two  expressions  for  the  velocity  are  true  whether  the 
particle  be  moving  in  a  right,  or  in  a  curved,  line. 

8.  Acceleration  is  the  rate  of  change  of  velocity.  It 
is  a  velocity  increment.  If  the  velocity  is  increasing,  the 
acceleration  is  considered  positive ;  if  decreasing,  it  is 
negative. 

Acceleration   is   said  to  be   uniform  when  the  velocity 


MEASURE   OF  ACCELERATION.  5 

receives  equal  increments  in  equal  times.     Otherwise  it  is 

variable. 

9.  Measure  of  Acceleration. — Uniform  acceleration 
is  measured  by  the  actual  increase  of  velocity  in  a  unit  of 
time.  Variable  acceleration  is  measured,  at  any  instant,  by 
the  velocity  which  would  be  generated  in  a  unit  of  time, 
were  the  velocity  to  increase,  during  that  unit,  at  the  same 
rate  as  at  the  instant  considered. 

Calling  /the  acceleration,  v  the  velocity,  and  t  the  time, 
we  have,  when  the  acceleration  is  uniform, 

/-r  PJ 

However  variable  the  acceleration  is,  it  may  be  regarded 
as  constant  during  the  infinitesimal  of  time  dt,  in  which 
time  the  increment  of  velocity  will  be  dv.  Hence,  denoting 
the  acceleration  at  the  time  t  by/,  we  have 

f=w  <*> 

We  also  have  (Art.  8) 

ds 

v  =  m 

which  in  (2)  gives 

f  _  dv  _    d      ds  _  d2s  .  . 

*  —  dt  ~  dt  '  Jt~~df  W 

That  is,  when  the  acceleration  is  variable  it  is  measured,  at 
any  instant,  by  the  derivative  of  the  velocity  regarded  as  a 
function  of  the  time,  or  by  the  second  derivative  of  the 
space  regarded  as  a  function  of  the  time. 
From  (3)  we  get,  by  integration,  when  /  is  constant, 

»  =  £—»  (4) 


6  VELOCITY  AND   ACCELERATION. 

ift%  =  s;  (5) 

and  2/5  =  v*}  (6) 

which  determine  the  velocity  and  space. 

10.  Geometric  Representation  of  Velocity  and 
Acceleration. — The  velocity  of  a  body  may  be  conveni- 
ently represented  geometrically  in  magnitude  and  direction 
by  means  of  a  straight  line.  Let  the  line  be  drawn  from 
the  point  at  which  the  motion  is  considered,  and  in  the 
direction  of  motion  at  that  point.  With  a  convenient  scale, 
let  a  length  of  the  line  be  cut  off  that  shall  contain  as  many 
units  of  length  as  there  are  units  in  the  velocity  to  be  repre- 
sented. The  direction  of  this  line  will  represent  the 
direction  of  the  motion,  and  its  length  will  represent  the 
velocity. 

Also  an  Ticceleratioit  may  be  represented  geometrically  by 
a  ^straight  line  drawn  in  the  direction  of  the  velocity 
generated,  and  containing  as  many  units  of  length  as  there 
are  units  of  acceleration  in  the  acceleration  considered. 
Also,  since  an  acceleration*  is  measured  by  the  actual 
increase  of  velocity  in  the  unit  of  time,  the  straight  line 
which  represents  an  acceleration  in  magnitude  and  direc- 
tion wjjl  also  completely  represent  the  velocity  generated  in 
theffnit  of  t«ne  to  which  the  acceleration  corresponds. 

11.  The  Mass  of  a  body  or  particle  is  the  quantity  of 
matter  which  it  contains;  and  is  proportional  to  the 
Volume  and  Density  jointly.  The  Density  may  therefore 
be  defined  as  the  quantity  of  matter  in  a  unit  of  volume. 

Let  M  be  the  mass,  p  the  density,  and  V  the  volume,  of 
a  homogeneous  body.    Then  we  have 

M  =  Vp,  (1) 

if  we  so  take  our  units  that  the  unit  of  mass  is  the  mass  of 
the  unit  volume  of  a  body  of  unit  density. 

*  Uniform  acceleration  is  here  meant. 


MOMENTUM.  7 

If  the  density  varies  from  point  to  point  of  the  body,  we 
have,  by  the  above  formula,  and  the  notation  of  the 
Integral  Calculus, 

M  =  fpd  V  =  fffpdx  cly  dz,  (2) 

where  p  is  supposed  to  be  a  known  function  of  #,  y,  z. 

In  England  the  unit  of  mass  is  the  imperial  standard 
pound  avoirdupois,  which  is  the  mass  of  a  certain  piece  of 
platinum  preserved  at  the  standard  office  in  London.  On 
the  continent  of  Europe  the  unit  of  mass  is  the  gramme. 
This  is  known  as  the  absolute  or  kinetic  unit  of  mass. 

12.  The  Quantity  of  Motion,*  o.^tlie  Momentum 
of  a  body  moving  without  rotation  is  the  product  of  its 
mass  and  velocity.  A  double  mass,  or  a  double  velocity, 
would  correspond  to  a  double  quantity  of  motion,  and 
so  on. 

Hence,  if  we  take  as  the  unit  of  momentum  the  mo- 
mentum of  the  unit  of  mass  moving  with  the  unit  of 
velocity,  the  momentum  of  a  mass  M  moving  with  velocity 

v  is  Mv.  . 

v«r 

13.  Change  of  Quantity  of  Motion,  or  Change  of 
Momentum,  is  proportional  to  the  mass  moving  and  the 
change  of  its  velocity  jointly.  If  then  the  mass  remains 
constant  the  change  of  momentum  is  measured  "by  the 
product  of  the  mass  into  the  change  of  velocity  ;  and  the 
rate  of  change  of  momentum,  or  acceleration  of  momentum, 
is  measured  by  the  product  of  the  mass  moving  and  the 
rate  of  change  of  velocity,  that  is,  by  the  product  of  the 
mass  moving  and  the  acceleration  (Art.  8).  Thus,  calling 
M the  mass,  we  have  for  the  measure  of  the  rate  of  change 
of  momentum, 

; *& 

*  This  phrase  was  used  by  Newton  in  place  of  the  more  modern  term  "  Momen- 
tum.11 


8  STATIC    MEASURE    OF   FORCE. 

14.  Force. — Force  is  any  cause  which  changes,  or  tends 
to  change,  a  tody's  state  of  vest  or  motion. 

A  force  always  tends  to  produce  motion,  but  may  be  pre- 
vented from  actually  producing  it  by  the  counteraction  of 
an  equal  and  opposite  force.  Several  forces  may  so  act  on 
a  body  as  to  neutralize  each  other.  When  a  body  remains 
at  rest,  though  acted  on  by  forces,  it  is  said  to  be  in 
equilibrium;  or,  in  other  words,  the  forces  are  said  to 
produce  equilibrium. 

What  force  is,  in  its  nature,  we  do  not  know.  Forces 
are  known  to  us  only  by  their  effects.  In  order  to  measure 
them  we  must  compare  the  effects  which  they  produce 
under  the  same  circumstances. 

15.  Static  Measure  of  Force. — The  effect  of  a  force 
depends  on:  1st,  its  magnitude,  or  intensity ;  2d,  its  direc- 
tion; i.  e.,  the  direction  in  which  it  tends  to  move  the  body 
on  which  it  acts  ;  and  3d,  its  point  of  application  ;  i.  e.,  the 
point  at  which  the  force  is  applied. 

The  effect  of  a  force  is  pressure,  and  may  be  expressed  by 
the  weight  which  will  counteract  it.  Every  force,  statically 
considered,  is  a  pressure,  and  hence  has  magnitude,  and 
may  be  measured.  A  force  may  produce  motion  or  not, 
according  as  the  body  on  which  it  acts  is  or  is  not  free  to 
move.  For  example,  take  the  case  of  a  body  resting  on  a 
table.  The  same  force  which  produces  pressure  on  the 
table  would  cause  the  body  to  fall  toward  the  earth  if  the 
table  were  removed. 

The  cause  of  this  pressure  or  motion  is  gravity,  or  the 
force  of  attraction  in  the  earth.  In  the  first  case  the  attrac- 
tion of  the  earth  produces  a  pressure;  in  the  second  case  it 
produces  motion.  Now  either  of  these,  viz.,  the  pressure 
which  the  body  exerts  when  at  rest,  or  the  quantity  of 
motion  it  acquires  in  a  unit  of  time,  may  be  taken  as  a 
means  of  measuring  the  magnitude  of  the  force  of  attrac- 
tion that  the  earth  exerts  on  the  body.     The  former  is 


METHOD    OF    COMPARING    FORCES.  9 

called  the  static  method,  and  the  forces  are  called  static 
forces;  the  latter  is  called  the  kinetic  method,  and  the 
forces  are  called  kinetic  forces.  Weight  is  the  name  given 
to  the  pressure  which  the  attraction  of  the  earth  causes  a 
body  to  exert.  Hence,  since  static  forces  produce  pressure, 
we  may  take,  as  the  unit  of  force,  a  pressure  of  one  pound 
(Art,  11). 

Therefore,  the  magnitude  of  a  force  may  he  measured 
statically  by  the  pressure  it  to  ill  produce  upo?i  some  body, 
and  expressed  in  pounds.  This  is  called  the  Static  measure 
of  force,  and  its  unit,  one  pound,  is  called  the  Gravitation 
unit  of  force. 

16.  Action  and  Reaction  are  always  equal  and 
opposite. — This  is  a  law  of  nature,  and  our  knowledge  of 
it  comes  from  experience.  If  a  force  act  on  a  body  held  by 
a  fixed  obstacle,  the  latter  will  oppose  an  equal  and  con- 
trary resistance.  If  the  force  act  on  a  body  free  to  move, 
motion  will  ensue ;  and,  in  the  act  of  moving,  the  inertia 
of  the  body  will  oppose  an  equal  and  contrary  resistance. 
If  we  press  a  stone  with  the  hand,  the  stone  presses  the 
hand  in  return.  If  we  strike  it,  we  receive  a  blow  by  the 
act  of  giving  one.  If  we  urge  it  so  as  to  give  it  motion,  we 
lose  some  of  the  motion  which  we  should  give  to  our  limbs 
by  the  same  effort,  if  the  stone  did  not  impede  them.  In 
each  of  these  cases  there  is  a  reaction  of  the  same  kind  as 
the  action,  and  equal  to  it. 

17.  Method  of  Comparing  Forces. — Two  forces  are 
equal  when  being  applied  in  opposite  directions  to  a 
particle  they  maintain  equilibrium.  If  we  take  two  equal 
forces,  and  apply  them  to  a  particle  in  the  same  direction, 
we  obtain  a  force  double  of  either  ;  if  we  unite  three  equal 
forces  we  obtain  a  triple  force  ;  and  so  on.  So  that,  in 
general,  to  compare  or  measure  forces,  we  have  only  to 
adopt  the  same  method  as  when  we  compare  or  measure 


10  REPRESENTATION    OF   FORCES. 

any  quantities  of  the  same  kind  ;  that  is,  we  must  take 
some  known  force  as  the  unit  of  force,  and  then  express,  in 
numbers,  the  relation  which  the  other  forces  bear  to  this 
•measuring  unit.  For  example,  if  one  pound  be  the  unit  of 
force  (Art.  15),  a  force  of  12  pounds  is  expressed  by  12; 
and  so  on. 

18.  Representation  of  Forces  by  Symbols  and 
Lines. — If  P.  Q.  Rv  etc.,  represent  forces,  they  are  numbers 
expressing  the  number  of  times  which  the  concrete  unit  of 
force  is  contained  in  the  given  forces. 

Forces  may  be  represented  geometrically  by  right  lines ; 
and  this  mode  of  representation  has  the  advantage  of  giving 
the  direction,  magnitude,  and  point  of  application  of  each 
force.  Thus,  draw  a  line  in  the  direction  of  the  given 
force ;  then,  having  selected  a  unit  of  length,  such  as  an 
inch,  a  foot,  etc.,  measure  on  this  line  as  many  units  of 
length  as  the  given  force  contains  units  of  weight.  The 
magnitude  of  the  force  is  represented  by  the  measured 
length  of  the  line ;  its  direction  by  the  direction  in  which 
the  line  is  drawn;  and  its  point  of  application  by  the  point 
from  which  the  line  is  drawn.* 

Thus,  let  the  force  P  act  at  the  point     * ? 

A,   in    the    direction   AB,   and    let    AB  F'9-  '• 

represent  as  many  units  of  length  as  P  contains  units  of 
force;  then  the  force  P  is  represented  geometrically  by 
the  line  AB ;  for  the  force  acts  in  the  direction  from  A 
to  B ;  its  point  of  application  is  at  A,  and  its  magnitude  is 
represented  by  the  length  of  the  line  AB. 

19.  Measure  of  Accelerating  Forces. — From  our 
definition  of  force  (Art.  14),  it  is  clear  that,  when  a  single 

*  Forces,  velocities,  anrl  accelerations  are  directed  quantities,  and  so  may  be 
represented  by  a  line,  in  direction  and  magnitude,  and  may  be  compounded  in  the 
same  way  as  vectors. 

If  anything  has  magnitude  and  direction,  the  magnitude  and  direction  taken 
together  constitute  a  vector: 


MEASURE    OF   ACCELERATING    FORCES.  11 

force  acts  upon  a  particle,  perfectly  free  to  move,  it  must 
produce  motion  :  and  hence  the  force  may  be  represented 
to  us  by  the  motion  it  lias  produced.  But  motion  is 
measured  in  terms  of  velocity  (Art.  6),  and  consequently  the 
velocity  communicated  to,  or  impressed  upon,  a  particle,  in 
a  given  time,  may  be  taken  as  a  measure  of  the  force. 
That  is,  if  the  same  particle  moves  along  a  right  line  so 
that  its  velocity  is  increased  at  a  constant  rate,  it  will  be 
acted,  upon  by  a  constant  force.  If  a  certain  constant  force, 
acting  for  a  second  on  a  given  particle,  generate  a  velocity 
of  32.2  feet  per  second,  a  double  force,  acting  for  one 
second  on  the  same  particle,  would  generate  a  velocity  of 
64.-i  feet  per  second ;  a  triple  force  would  generate  a 
velocity  of  96.6  feet  per  second,  and  so  on. 

If  the  rate  of  increase  of  the  velocity,  (t.  e.,  the  accelera- 
tion), of  the  particle  is  not  uniform,  the  force  acting  on  it 
is  not  uniform,  and  the  magnitude  of  the  force,  at  any 
point  of  the  particle's  path,  is  measured  by  the  acceleration 
of  the  particle  at  this  point.  Hence,  since  one  and  the 
same  particle  is  capable  of  moving  with  all  possible  accelera- 
tions, all  forces  may  be  measured  by  the  velocities  they 
generate  in  the  same  or  equal  particles  in  the  same  or  equal 
times.  When  forces  are  so  measured  they  are  called 
Accelerating  Forces. 

20.  Kinetic  or  Absolute  Measure  of  Force.* — Let 

n  equal  particles  be  placed  side  by  side,  and  let  each  of  them 
be  acted  on  uniformly  for  the  same  time,  by  the  same  force. 
Each  particle,  at  the  end  of  this  time,  will  have  the  same 
velocity.  Now  if  these  n  separate  particles  are  all  united  so 
as  to  form  a  body  of  n  times  the  mass  of  each  particle,  and 
if  each  one  of  them  is  still  acted  on  by  the  same  force  as 

*  Arts.  20,  21,  22,  and  25,  treat  of  the  Kinetic  measure  of  force,  and  may  be 
omitted  till  Part  III  is  reached  ;  but  it  is  convenient  to  present  them  once  for  all, 
and,  for  tne  sake  of  reference  and  comparison,  to  place  them  with  the  Static 
measure  of  force  at  the  beginning  of  the  work. 


12        KINETIC    OR    ABSOLUTE    MEASURE    OF   FORCE. 

before,  this  body,  at  the  end  of  the  time  considered,  will 
have  the  same  velocity  that  each  separate  particle  had,  and 
will  be  acted  on  by  n  times  the  force  which  generated  this 
velocity  in  the  particle.  Comparing  a  single  particle,  then, 
with  the  body  whose  mass  is  n  times  the  mass  of  this 
particle,  we  see  that,  to  produce  the  same  velocity  in  two 
bodies  by  forces  acting  on  them  for  the  same  time,  the 
magnitudes  of  the  forces  must  be  proportional  to  the 
masses  on  which  they  act.*  Hence,  generally,  since  force 
varies  as  the  velocity  when  the  mass  is  constant  (Art.  19), 
and  varies  as  the  mass  when  the  velocity  is  constant,  we 
have,  by  the  ordinary  law  of  proportion,  when  both  are 
changed,  force  varies  as  the  product  of  the  mass  acted  upon 
and  the  velocity  generated  in  a  given  time  ;  that  is,  it  varies 
as  the  quantity  of  motion  (Art.  13)  it  produces  in  a  given 
mass  in  a  given  time.  If  the  force  be  variable,  the  rate  of 
change  of  velocity  is  variable  (Art.  19),  and  hence  the  force 
varies  as  the  product  of  the  mass  on  which  it  acts  and  the 
rate  of  change  of  velocity,  i.  e.,  it  varies  as  the  acceleration 
of  the  momentum  (Art.  13).  Therefore,  if  any  force  P  act 
on  a  mass  M,  we  have 

P  «  Mf;  (1) 

or,  in  the  form  of  an  equation 

P  =  kMf>  (8) 

where  h  is  some  constant. 

If  the  unit  of  force  be  taken  as  that  force  which,  acting 
on  the  unit  of  mass  for  the  unit  of  time,  generates  the  unit 
of  velocity,  then  if  we  put  M  equal  to  unity,  i.  e.,  take  the 
unit  of  mass,  and  /equal  to  unity,  i.  e.,  take  the  unit  of 
acceleration,  we  must  have  the  force  producing  the  accel- 
eration equal  to  the  unit  of  force,  or  P  equal  to   unity. 

*  Minchin's  Statics,  p.  5. 


THE  ABSOLUTE    OR   KINETIC  MEASURE   OF  FORCE.    13 

Hence  h  must  also  be  equal  to  unity,  and  we  have  the 
equation, 

P  =  Mf.  (3) 

Therefore,  the  Kinetic  or  Absolute  measure  of  a  force  is 
the  rate  of  change  or  acceleration*  of  momentum  it  produces 
in  a  unit  of  time. 

If  the  force  is  constant,  (3)  becomes  by  (1)  of  Art.  9, 

P  =  f  •  (4) 

And  if  the  force  is  variable,  (3)  becomes  by  (3)  of  Art.  9, 
P  =  M§  (5) 

21.  The  Absolute   or  Kinetic   Unit  of  Force.— 

A  second,  a  foot,  and  a  pound  being  the  units  of  time,  space, 
and  mass,  respectively  (Arts.  6  and  11),  we  are  required  to 
find  the  corresponding  unit  of  force  that  the  above  equation 
may  be  true.  The  unit  of  force  is  that  force  which,  acting 
for  one  second,  on  the  mass  of  one  pound,  generates  in  it  a 
velocity  of  one  foot  per  second.  Now,  from  the  results  of 
numerous  experiments,  it  has  been  ascertained  that  if  a 
body,  weighing  one  pound,  fall  freely  for  one  second  at  the 
sea  level,  it  will  acquire  a  velocity  of  about  32.2  feet  per 
second  ;  i.  e.,  a  force  equal  to  the  weight  of  a  pound,  if 
acting  on  the  mass  of  a  pound,  at  the  sea  level,  generates  in 
it  in  one  second,  if  free  to  move,  a  velocity  of  nearly  32.2 
feet  per   second.      It   follows,  therefore,  that  a  force  of 

^—=  of  the  weight  of  a  pound,  if  acting  on  the  mass  of 

a  pound,  at  the-  sea  level,  generates  in  it  in  one  second,  if 
free  to  move,  a  velocity  of  one  foot  per  second  ;  and  hence 

*  See  Tait  and  Steele's  Dynamics  of  a  Particle,  p.  43. 


14  MEASURES   OF  FORCE. 

the  unit  of  force  is  — -  of  the  weight  of  a  pound,  or  rather 

less  than  the  weight  of  half  an  ounce  avoirdupois  ;  so  that 
half  an  ounce,  acting  on  the  mass  of  a  pound  for  one 
second,  will  give  to  it  a  velocity  of  one  foot  per  second. 
This  is  the  British  absolute  kinetic*  unit  of  force. 

In  order  that  Eq.  3  (Art.  20)  may  be  universally  true 
when  a  second,  a  foot,  and  a  pound  are  the  units  of  time, 
space,  and  mass  respectively,  all  forces  must  be  expressed  in 
terms  of  this  unit. 

22.  Three  Ways  of  Measuring  Force. — (1.)  If  a 
force  does  not  produce  motion  it  is  measured  by  the  pres- 
sure it  produces,  or  the  number  of  pounds  it  will  support 
(Ar.t.  15).  This  is  the  measure  of  Static  Force,  and  its 
unit  is  the  weight  of  a  pound. 

(2.)  If  we  consider  forces  as  always  acting  on  a  unit  of 
mass,  and  suppose  that  there  are  no  forces  acting  in  the 
opposite  direction,  then  these  forces  will  be  measured 
simply  by  the  velocities  or  accelerations  which  they  generate 
in  a  given  time.  This  is  the  measure  of  Accelerating  Force, 
and  its  unit  is  that  force  ivhich,  acting  on  the  unit  of  mass, 
during  the  unit  of  tune,  generate  the  unit  of  velocity ; 
hence  (Art.  21),  the  unit  of  force  is  the  force  ivhicfr,  acting 
on  one  pound  of  mass  for  one  second,  generates  a  velocity  of 
one  foot  per  second. 

(3.)  If  forces  act  on  different  masses,  and  produce  motion 
in  them,  and  we  consider  as  before  that  there  are  no  forces 
acting  in  the  opposite  direction,  then  the  forces  are  meas- 
ured by  the  quantity  of  motion,  or  by  the  acceleration  of 
momentum  generated  in  a  unit  of  time  (Art.  20).  This  is 
the  measure  of  Moving  Force,  and  its  unit  (Art.  21)  is  the 
force  which,  acting  on  one  pound  of  mass  for  one  second, 
generates  a  velocity  of  one  foot  per  second. 

*  Introduced  by  Gauss. 


MEANING    OF  G    IN  DYNAMICS.  15 

It  must  be  understood  thai  when  we  speak  of  static, 
accelerating,  or  moving  forces,  we  do  not  refer  bo  differept 
kinds  of  force,  but  only  to  force  as  measured  in  different 
ways. 

23.  Meaning  of  g  in  Dynamics. — The  most  impor- 
tant case  of  a  constant,  or  very  nearly  constant,  force  is 
gravity  at  the  surface  of  the  earth.  The  force  of  gravity  is 
so  nearly  constant  for  places  near  the  earth's  surface,  thai 
falling  bodies  may  be  taken  as  examples  of  motion  under  a 
constant  force.  A  stone,  let  fall  from  rest,  moves  at  first 
very  slowly.  During  the  first  tenth  of  a  second  the  velocity 
is  very  small.  In  one  second  the  stone  has  acquired  a 
velocity  of  about  32  feet  per  second. 

A  great  number  of  experiments  have  been  made  to  ascer- 
tain the  exact  velocity  which  a  body  would  acquire  in  one 
second  under  the  action  of  gravity,  and  freed  from  the 
resistance  of  the  air.  The  most  accurate  method  is  indi- 
rect, by  means  of  the  pendulum.  The  result  of  pendulum 
experiments  made  at  Leith  Fort,  by  Captain  Kater,  is, 
that  the  velocity  acquired  by  a  body  falling  unresisted  for 
one  second  is,  at  that  place,  32.207  feet  per  second.  The 
velocity  acquired  in  one  second,  or  the  acceleration  (Art. 
8),  of  a  body  falling  freely  in  vacuo,  is  found  to  vary 
slightly  with  the  latitude,  and  also  with  the  elevation  above 
the  sea  level.  In  London  it  is  32.1889  feet  per  second.  In 
latitude  45°,  near  Bordeaux,  it  is  32.1703  feet  per  second. 

This  acceleration  is  usually  denoted  by  g ;  and  when  we 
say  that  at  any  place  g  is  equal  to  32,  we  mean  that  the 
velocity  generated  per  second  in  a  body  falling  freely* 
under  the  action  of  gravity  at  that  place,  is  a  velocity  of 
32  feet  per  second.  The  average  value  of  g  for  the  whole 
of  Great  Britain  differs  but  little  from  32.2  ;  and  hence  the 
numerical  value  of  g  for  that  country  is  taken  to  be  32.2. 

*  A  body  is  said  to  be  moving  freely  when  it  is  acted  upon  by  no  forces  except 
those  under  consideration. 


16  THE    UNIT  OF  MASS. 

The  formula,  deduced  from  observation,  and  a  certain 
theory  regarding  the  figure  and  density  of  the  earth,  which 
may  be  employed  to  calculate  the  most  probable  value  of 
the  apparent  force  of  gravity,  is 

g  =  G(l  +  .005133  sin2  X), 

where  G  is  the  apparent  force  of  gravity  on  a  unit  mass  at 
the  equator,  and  g  the  force  of  gravity  in  any  latitude  A; 
the  value  of  G,  in  terms  of  the  British  absolute  unit,  being 
32.088.     (See  Thomson  and  Tait,  p.  226.) 

24.  Gravitation  Units  of  Force  and  Mass. — If  in 

(3)  of  Art.  20,  we  put  for  P,  the  weight  W  of  the  body, 
and  write  g  for  /  since  we  know  the  acceleration  is  g,  (3) 

becomes 

W  =  mg.  (1) 

W 
.-.    i»  =  -.  (2) 

9 

W 
and  hence  —  may  be  taken  as  the  measure  of  the  mass. 

In  gravitation  measure  forces  are  measured  by  the  pres- 
sure they  tvill  produce,  and  the  unit  of  force  is  one  pound 
(Art.  15),  and  the  unit  of  mass  is  the  quantity  of  matter  in 
a  body  which  weighs  g  pounds  at  that  place  where  the  accel- 
eration of  gravity  is  g. 

This  definition  gives  a  unit  of  mass  which  is  constant  at 
the  same  place,  but  changes  with  the  locality;  i.  e.,  its  iveight 
changes  with  the  locality  while  the  quantity  of  matter  in  it 
remains  the  same.  Thus,  the  unit  of  mass  would  weigh  at 
Bordeaux  32.1703  pounds  (Art.  23),  while  at  Leith  Fort  it 
would  weigh  32.207  pounds.  Let  m  be  the  mass  of  a  body 
which  weighs  w  pounds.  The  quantity  of  matter  in  this 
body  remains  the  same  when  carried  from  place  to  place. 
If  it  were  possible  to  transport  it  to  another  planet  its  mass 


GRAVITATION  MEASURE   OF  FORCE,  17 

would  not  bo  altered,  but  its  weight  would  be  very  different, 
Its  weight  wherever  placed  would  vary  directly  as  the  force 
of  gravity  ;  but  the  acceleration  also  would  vary  directly  as 
the  force  of  gravity.  If  placed  on  the  sun,  for  example,  it 
would  weigh  about  28  times  as  much  as  on  the  surface  of 
the  earth  ;  but  the  acceleration  on  the  sun  would  also  be 
28  times  as  much  as  on  the  surface  of  the  earth  ;  that  is, 
the  ratio  of  the  wreight  to  the  acceleration,  anywhere  in 

the  universe  is   constant,    and  hence    — ,  which    is  the 

9 
numerical  value  of  m  (Eq.  2),  is  constant  for  the  same 
mass  at  all  places. 

25.  Comparison  of  Gravitation  and  Absolute 
Measure. — The  pound  weight  has  been  long  used  for  the 
measurement  of  force  instead  of  mass,  and  is  tbe  recognized 
standard  of  reference.  It  came  into  general  use  because  it 
afforded  the  most  ready  and  simple  method  of  estimating 
forces.  The  pressure  of  steam  in  a  boiler  is  always  reck- 
oned in  pounds  per  square  inch.  The  tension  of  a  string  is 
estimated  in  pounds;  the  force  necessary  to  draw  a  train  of 
cars,  or  the  pressure  of  water  against  a  lock-gate,  is 
expressed  in  pounds.  Such  expressions  as  "a  force  of 
10  pounds,"  or  "  a  pressure  of  steam  equal  to  50  pounds  on 
the  inch,"  are  of  every  day  occurrence.  Therefore  this 
method  of  measuring  forces  is  eminently  convenient  in 
practice.  For  this  reason,  and  because  it  is  the  one  used 
by  most  engineers  and  writers  of  mechanics,  we  shall  adopt 
it  in  this  work,  and  adhere  to  the  measurement  of  force  by 
pounds,  and  give  all  our  results  in  the  usual  gravitation 
measure.     In  this  measure  it  is  convenient  to  represent  the 

W 
mass  of  a  body   weighing   W  pounds  by  the  fraction  — 

(Art.  24:),  so  that  (3)  of  Art.  20  becomes 

gj  \ 


18  EXAMPLES. 

To  do  so  it  will  only  be  necessary  to  assume  that  the  unit 
of  mass  is  the  quantity  of  matter  in  a  body  weighing  g 
pounds,  and  changes  in  weight  in  the  same  proportion  that 
g  changes  (Art.  24). 

Of  course,  the  units  of  mass  and  force  in  (3)  of  Art.  20 
may  be  either  absolute  or  gravitation  units.  If  absolute, 
the  unit  of  mass  is  one  pound  (Art.  11),  and  the  unit  of 

force  is  -  pounds  (Art.  21).     If  gravitation,  the  units  are 

g  times  as  great;  L  e.,  the  unit  of  mass  is  g  pounds  (Art. 
24),  and  the  unit  of  force  is  one  pound  (Art.  15). 

The  advantage  of  the  gravitation  measure  is,  it  enables  us 
to  express  the  force  in  pounds,  and  furnishes  us  with  a  con- 
stant numerical  representative  for  the  same  quantity  of 
matter;  that  is  to  say,  a  mass  represented  by  20  on  the 
equator  would  be  represented  by  20,  at  the  pole  or  on 
the  sun.  Hence,  in  (1),  P  is  the  static  measure  of 
any  moving  force  [Art.  22,  (3)],  W  is  the  weight  of  the 
body  in   pounds,  g  the  acceleration  of  gravity  (Art.  23), 

W 

—  the  mass  upon  which  the  force  acts  [(2)  of  Art.  24],  and 

which  is  free  to  move  under  the  action  of  P,  the  unit  of 
mass  being  the  mass  weighing  g  pounds,  and  /  the 
acceleration  which  the  force  P  produces  in  the  mass. 

EXAMPLES 

1.  Compare  the  velocities  of  two  points  which  move 
uniformly,  one  through  5  feet  in  half  a  second,  and  the 
other  through  100  yards  in  a  minute.    Ans.  As  2  is  to  1. 

2.  Compare  the  velocities  of  two  points  which  move  uni- 
formly, one  through  720  feet  in  one  minute,  and  the  other 
through  3 J  yards  in  three-quarters  of  a  second. 

An*.  As  6  is  to  7. 
6.  A  railway  train  travels  100  miles  in   2  hours  ,   *ind 
the  average  velocity  in  feet  per  second.  Ans.   73^. 


EXAMPLES.  19 

4.  One  point  moves  uniformly  round  the  circumference 
of  a  circle,  while  another  point  moves  uniformly  along 
the  diameter  ;  compare  their  velocities. 

Ans.  As  77  is  to  1. 

5.  Supposing  the  earth  to  be  a  sphere  25000  miles  in 
circumference,  and  turning  round  once  in  a  day,  deter- 
mine the  velocity  of  a  point  at  the  equator. 

Ans,  1527-$  ft.  per  sec. 

G.  A  body  has  described  50  feet  from  rest  in  2  seconds, 
with  uniform  acceleration  ;  find  the  velocity  acquired. 

From  (5)  of  Art.  9  we  have 

/=25; 

and  from  (4)  we  have      ft  =  v; 

.'.     v  =  50. 

7.  Find  the  time  it  will  take  the  body  in  the  last  exam- 
ple to  move  over  the  next  150  feet. 

From  (5)  of  Art.  9  we  have 

8  =  yp i    .'.    etc. 

Ans,  2  seconds. 

8.  A  body,  moving  with  uniform  acceleration,  describes 
63  feet  in  the  fourth  second  ;  find  the  acceleration. 

Ans.  18. 

9.  A  body,  with  uniform  acceleration,  describes  72  feet 
while  its  velocity  increases  from  16  to  20  feet  per  second; 
find  the  whole  time  of  motion,  and  the  acceleration. 

Ans.  20  seconds ;  1. 

10.  A  body,  in  passing  over  9  feet  with  uniform  accelera- 
tion, has  its  velocity  increased  from  1  to  5  feet  per  second; 
find  the  whole  space  described  from  rest,  and  the  accelera- 
tion. Ans.  25  feet ;  \, 


20  EXAMPLES. 

11.  A  body,  uniformly  accelerated,  is  found  to  be  mov- 
ing at  the  end  of  10  seconds  with  a  velocity  which,  if 
continued  uniformly,  would  carry  it  through  45  miles  in 
the  next  hour  ;  find  the  acceleration.  Ans.  6f. 

12.  Find  the  mass  of  a  straight  wire  or  rod,  the  density 
of  which  varies  directly  as  the  distance  from  one  end. 

Take  the  end  of  the  rod  as  origin  ;  let  a  =  its  length  ; 
let  the  distance  of  any  point  of  it  from  that  end  =  x ; 
and  let  w  =  the  area  of  its  transverse  section,  and  k  =  the 
density  at  the  unit's  distance  from  the  origin.     Then 

d  V  =  o)dx ;    and    p  =  Tex ; 

and  (2)  of  Art.  11  becomes 


M  =    I     k(ox  dx  = 


13.  Find  the  mass  of  a  circular  plate  of  uniform  thick- 
ness, the  density  of  which  varies  as  the  distance  from  the 
centre. 

Ans.  §7rMa3,  where  a  is  the  radius,  h  the  density  at 
the  unit's  distance,  and  h  the  thickness. 

14.  Find  the  mass  of  a  sphere,  whose  density  varies 
inversely  as  the  distance  from,  the  centre. 

Ans.  2npa?i  where  p  is  the  density  of  the  outside  stratum. 


STATICS     (REST) 


CHAPTER     II. 

THE     COMPOSITION     AND     RESOLUTION     OF    CONCUR. 
RING     FORCES— CONDITIONS     OF     EQUILIBRIUM. 

26.  Problem  of  Statics. —The  primary  conception  of 
force  is  that  of  a  cause  of  motion  (Art.  14).  If  only  one 
force  acts  on  a  particle  it  is  clear  that  the  particle  cannot 
remain  at  rest.  In  statics  it  is  only  the  tendency  which^ 
forces  have  to  produce  motion  that  is  considered.  There 
must  be  at  least  two  forces  in  statics ;  and  they  are  con- 
sidered as  acting  so  as  to  counteract  each  other's  tendency 
to  cause  motion,  thereby  producing  a  state  of  equilibrium 
in  the  bodies  to  which  they  are  applied.  The  forces  which 
act  upon  a  body  may  be  in  equilibrium,  and  yet  motion 
exist;  but  in  such  cases  the  motion  is  uniform.  Hence 
there  are  two  kinds  of  equilibrium,  the  one  relatiug  to 
bodies  at  rest,  the  other  relating  to  bodies  in  motion.  The 
former  is  sometimes  called  Static  Equilibrium  and  the  lat- 
ter Kinetic  (or  Dynamic*)  Equilibrium.  The  problem  of 
statics  is  to  determine  the  conditions  under  'which  forces  act 
when  they  keep  bodies  at  rest. 

27.  Concurring  and  Conspiring  Forces. — Result- 
ant.— When  several  forces  have  a  common  point  of  appli- 
cation they  are  called  concurring  forces  ;  when  they  act  at 
the  same  point  and  along  the  same  right  line  they  are 
called  conspiring  forces. 

The  resultant  of  two  or  more  forces  is  that  force  wdiich 
singly  will  produce  the  same  effect  as  the  forces  them- 
selves when  acting  together.  The  individual  forces,  when 
considered  with   reference   to    this    resultant,    are    called 

*  Gregory's  Mechanics,  p.  14. 


22  COMPOSITION   OF  CONSPIRING   FORCES. 

components.     The  process  of  finding  the  resultant  of  several 
forces  is  called  the  compositioji  of  forces. 

28.  Composition  of  Conspiring  Forces. — Condi- 
tion of  Equilibrium. — When  two  or  more  conspiring 
forces  act  in  the  same  direction,  it  is  evident  that  the 
resultant  force  is  equal  to  their  sum,  and  acts  in  the  same 
direction. 

When  two  conspiring  forces  act  in  opposite  directions 
their  resultant  force  is  equal  to  their  difference,  and  acts  in 
the  direction  of  the  greater  component. 

When  several  conspiring  forces  act  in  different  directions 
the  resultant  of  the.  forces  acting  in  one  direction  equals 
the  sum  of  these  forces,  and  acts  in  the  same  direction  ; 
and  so  of  the  forces  acting  in  the  opposite  direction. 
Therefore,  the  resultant  of  all  the  forces  is  equal  to  the 
difference  of  these  sums,  and  acts  in  the  direction  of  the 
greater  sum.  Hence,  if  the  forces  acting  in  one  direction 
are  reckoned  positive,  and  those  in  the  opposite  direction 
negative,  their  resultant  is  equal  to  their  algebraic  sum  ; 
its  sign  determining  the  direction  in  which  it  acts.  Thus, 
if  Pt,  P2,  P3,  etc.,  are  the  conspiring  forces,  some  of 
which  may  be  positive  and  the  others  negative,  and  R  is 
the  resultant,  we  have 

R  =  Pt  +  P2  +  P3  +  etc.  =  ZP,  (1) 

in  which  2  denotes  the  algebraic  sum  of  the  terms  similar 
to  that  written  immediately  after  it. 

Cor. —The  condition  that  the  forces  may  be  in  equilib- 
rium is  that  their  resultant,  and  therefore  their  algebraic 
sum,  must  vanish.  Hence,  when  the  forces  are  in  equilib- 
rium we  must  have  R  =  0  ;  therefore  (1)  becomes 

Px  +  P2  +  P3  +  etc.  =  LP  =  0.  (2) 


COMPOSITION    OF    VELOCITIES.  23 

29.  Composition  of  Velocities.—//^  particle  be 
moving  with  two  uniform  velocities  represented  in, 

magnitude  and  direction  by  the  two  adjacent  sides 
of  a  parallelogram,  the  resultant  velocity  will  be 
represented  in  magnitude  and  direction  by  the 
diagonal  of  the  parallelogram. 

Let  the  particle  move  with  a  uniform 
velocity  v,  which  acting  alone  will  take 
it  in  one  second  from  A  to  B,  and  with 
a  uniform  velocity  v' ,  which  acting 
alone  will  take  it  in  one  second  from  A 
to  C  ;  at  the  end  of  one  second  the  par- 
ticle will  be  found  at  D,  and  AD  will  represent  in  magni- 
tude and  direction  the  resultant  of  the  velocities  represented 
by  AB  and  AC. 

Suppose  the  particle  to  move  uniformly  along  a  straight 
tube  which  starts  from  AB,  and  moves  uniformly  parallel 
to  itself  with  its  extremity  in  AC.  When  the  particle  starts 
from  A  the  tube  is  in  the  position  AB.  When  the  particle 
has  moved  over  any  part  of  AB,  the  end  of  the  tube  has 
moved  over  the  same  part  of  AC,  and  the  particle  is  on  the 

line  AD.     For  example,  let  AM  be  the  -th  part  of  AB,  and 

AN  be  the  -th  part  of  AC  ;  while  the  particle  moves  from 

A  to  M,  the  end  A  with  the  tube  AB  will  move  from  A  to 
N,  and  the  particle  will  be  at  P,  the  tube  occupying  the 
position  NL,  and  PM  being  parallel  and  equal  to  AN.  P 
can  be  proved  to  be  on  the  diagonal  AD  as  follows  : 

AT)  AH 

AM  :  MP    ::    —  :  —     ::    AB  :  AC  (=  BD); 
n         n  . 

therefore  P  lies  on  the  diagonal  AD.    Also  since 

AM  :  AB    : :    AP  :  AD, 


24  COMPOSITION  OF  FORCES. 

the  resultant  velocity  is  uniform.  Hence,  the  diagonal  AD 
represents  in  magnitude  and  direction  the  resultant  of  the 
velocities  represented  by  AB  and  AC. 

This  proposition  is  known  as  the  Parallelogram  of 
Velocities. 

30.  Composition  of  Forces.— From  the  Parallelo- 
gram of  Velocities  the  Parallelogram  of  Forces  folio ws 
immediately.  Since  two  simultaneous  velocities,  AB  and 
AC,  of  a  particle,  result  in  a  single  velocity,  AD,  and  since 
these  three  velocities  may  be  regarded  as  the  measures  of 
three  separate  forces  all  acting  for  the  same  time  (Art.  19), 
it  follows  that  the  effect  produced  on  a  particle  by  the  com- 
bined action,  for  the  same  time,  of  two  forces  may  be  pro- 
duced by  the  action,  for  the  same  time,  of  a  single  force, 
which  is  therefore  called  the  resultant  of  the  other  two 
forces ;  and  these  forces  are  represented  in  magnitude  and 
direction  by  AB,  AC,  and  AD.  (See  Minchin,  p.  7,  also 
Garnett's  Dynamics,  p.  10.) 

Hence  if  two  concurring  forces  be  represented  in  magni- 
tude and  direction  by  the  adjacent  sides  of  a  parallelogram, 
their  resultant  will  be  represented  in  magnitude  and  direction 
by  the  diagonal  of  the  parallelogram.  Care  must  be  taken 
in  constructing  the  parallelogram  of  forces  that  the  com- 
ponents both  act  from  the  angle  of  the  parallelogram  from 
which  the  diagonal  is  drawn. 

This  proposition  has  been  proved  in  various  ways.  It  was  enun- 
ciated in  its  present  form  by  Sir  Isaac  Newton,  and  by  Varignon,  the 
celebrated  mathematician,  in  the  year  1687,  probably  independent  of 
each  other.  Since  that  time  various  proofs  of  it  have  been  given  by 
different  mathematicians.  One  work  gives  a  discussion,  more  or  less 
complete,  of  45  other  proofs.  A  noted  analytic  proof  is  given  by 
M.  Poisson.  (See  Price's  Cal.,  Vol.  Ill,  p.  19).  Some  authors  object 
to  proving  the  parallelogram  of  forces  by  means  of  the  parallelogram 
of  velocities.  (See  Gregory's  Mechanics,  p.  14.)  The  student  who 
wants  other  proofs  is  referred  to  Duchayla's  proof  as  found  in  Tod- 
hunter's  Statics,  p.  7,  and  in  Galbraith's  Mechanics,  p,  7,  and  in  many 


TRIAXGLE   OF  FORCES.  25 

other  works  ;  or  to  Laplace's  proof.  (See  Mecanique  Celeste,  Liv.  I, 
chap.  1.) 

If  S  be  the  angle  between  the  sides  of  the  parallelogram, 
AB  and  AC  (Fig.  2),  and  P  and  Q  represent  the  two  com- 
ponent forces  acting  at  A,  and  R  represent  the  resultant, 
AD,  we  have  from  trigonometry, 

R?  =  P2  +  Q2  +  2PQ  cos  0  (1) 

an  equation  which  gives  the  magnitude  of  the  resultant  of 
two  forces  in  terms  of  the  magnitudes  of  the  two  forces  and 
the  angle  between  their  directions,  the  forces  being  repre- 
sented by  two  lines,  both  drawn  from  the  point  at  which 
they  act. 

Cor. — If  6  =  90°,  and  «  and  13  be  the  angles  which  the 
direction  of  R  makes  with  the  directions  of  P  and  Q,  we 
have  from  (1) 

(2) 

p  \ 
Also 

(3) 


from  which  the  magnitude  and  direction  of  the  resultant 
are  determined. 

31.  Triangle  of  Forces. — //  three  concurring 
forces  be  represented  in  magnitude  and  direction 
by  the  sides  of  a  triangle,  tahen  in  order,  they  will 
be  in  equilibrium. 

Let  ABC    be   the    triangle  whose 
sides,    taken   in   order,    represent   in 
magnitude  and  direction  three  forces 
applied  at   the   point   A.     Complete     o        pjfl,3 
2 


R2  = 

=  P2  +  Q' 

cos  a  = 

P  \ 

~  R'l 

cos  j3  = 

"  R9) 

26  TRIANGLE   OF  FORCES. 

the  parallelogram  ABCD.  Then  the  forces,  AB  and  BC, 
applied  at  A,  are  expressed  by  AB  and  AD  (since  AD  is 
equal  and  parallel  to  BC).  But  the  resultant  of  AB  and 
AD  is  AC,  acting  in  the  direction  AC.  Therefore  the  three 
forces  represented  by  AB,  BC,  and  CA,  are  equivalent  to 
two  forces,  AC  and  CA,  the  former  acting  from  A  towards 
C  and  the  latter  from  C  towards  A,  which,  being  equal  and 
opposite,  will  clearly  balance  each  other.  Therefore  the 
three  forces  represented  by  AB,  BC,  and  CA,  acting  at  the 
point  A,  will  be  in  equilibrium. 

It  should  be  observed  that  though  BC  represents  the 
magnitude  and  direction  of  the  component,  it  is  not  in  the 
line  of  its  action,  because  the  three  forces  act  at  the 
point  A. 

The  converse  of  this  is  also  true  ;  viz.,  If  three  concurring 
forces  are  in  equilibrium,  they  may  be  represented  in  mag- 
nitude and  direction  by  the  sides  of  a  triangle,  drawn 
parallel  respectively  to  the  directions  of  the  forces. 

Thus,  if  AB  and  BC  represent  two  forces  in  magnitude 
and  direction,  AC  will  represent  the  resultant,  and  hence  to 
produce  equilibrium  the  resultant  force  AC  must  be  opposed 
by  an  equal  and  opposite  force  CA.  Therefore,  the  three 
forces  in  equilibrium  will  be  represented  by  AB,  BC,  and 
CA. 

Cor. — When  three  concurring  forces  are  in  equilibrium, 
each  is  equal  and  directly  opposite  to  the  resultant  of  the 
other  two. 

32.  Relations  between  Three  Concurring  Forces 
in  Equilibrium. — Since  the  sides  of  a  plane  triangle  are 
as  the  sines  of  the  opposite  angles,  we  have  (Fig.  3) 

AB  :  BC  (or  AD)  :  AC   : :   sin  ACB  :  sin  BAC  :  sin  ABC 
: :    sin  DAC  :  sin  BAC  :  sin  BAD. 

Hence,  calling  P,  Q,  and  i?,  the  forces  represented  by  AB, 
AD,  and  AC,  and  denoting  the  angles  between  the  direc- 


POLYGON    OF    FORCES.  21 

tions  of  the  forces  P  and  Q,  Q  and  R,  and  R  and  P,  by 

A      A  A 

PQ,  QR,  and  PP,  respectively,  we  have 

_*_  =  _«_  =  La_         (1) 

A  A  A  v  ' 

sin  §P        sin  RP        sin  P() 

Therefore,  when  three  concurring  forces  are  in  equilibrium 
they  are  respect i rely  in  the  same  proportion  as  the  sines  of 
the  angles  included  between  the  directions  of  the  other  two. 

33.  The  Polygon  of  Forces.—  //'  any  number  of 
concurring  forces  be  represented,  in  magnitude  and 
direction  by  the  sides  of  a  closed  polygon  taken  in 
order,  they  will  be  in  equilibrium. 

Let  the  forces  be  represented  in 
magnitude  and  direction  by  the  lines 
AP1?  AP25  AP3,  AP4,  AP5.  Take 
AB  to  represent  AP1?  through  B  draw 
BC  equal  and  parallel  to  AP2  ;  the 
resultant  of  the  forces  AB  and  BC,  or 
APX  and  AP2  is  represented  by  AC 
(Art.   31).      Of  course  the  force,  BO,  Fi9'4 

acts  at  A  and  is  parallel  to  BC.  Again  through  0  draw  OD 
equal  and  parallel  to  AP3,  the  resultant  of  AC  and  CD,  or 
APj,  AP2,  and  AP3  is  AD.  Also  through  D  draw  DE 
equal  and  parallel  to  AP4,  the  resultant  of  AD  and  DE,  or 
AP17  AP2,  AP3,  and  AP4  is  AE.  Now  if  AE  is  equal  and 
opposite  to  AP5  the  system  is  in  equilibrium  (Art.  18). 
Hence  the  forces  represented  by  AB,  BC,  CD,  DE,  EA 
will  be  in  equilibrium. 

Cor.  1. — Any  one  side  of  the  polygon  represents  in 
magnitude  and  direction  the  resultant  of  all  the  forces 
represented  by  the  remaining  sides. 

Cor.  2. — If  the  lines  representing  the  forces  do  not  form 
n  closed  polygon  the  forces  are  not  in  equilibrium  ;  in  this 


28 


PARALLELOPIPED     OF   FORCES, 


case  the  last  side,  AE,  taken  from  A  to  E,  or  that  which  is 
required  to  close  up  the  polygon,  represents  in  magnitude 
and  direction  the  resultant  of  the  system. 

34.  Parallelopiped  of  Forces.—  If  three  concur- 
ring forces,  not  in  the  same  plane,  are  represented 
in  magnitude  and  direction  by  the  three  edges  of 
a  parallelopiped,  then  the  resultant  will  be  repre- 
sented in  magnitude  and  direction  by  the  diag- 
onal; conversely,  if  the  diagonal  of  a  parallel- 
opiped represents  a  force,  it  is  equivalent  to  three 
forces  represented  by  the  edges  of  the  parallel- 
opiped. 

Let  the  three  edges  AB,  AC,  AD  of  the 
parallelopiped  represent  the  three  forces, 
applied  at  A.  Then  the  resultant  of  the 
forces  AB  and  AC  is  AE,  the  diagonal  of 
the  face  ABOE;  and  the  resultant  of  the 
forces  AE  and  AD  is  AF,  the  diagonal  of 
the  parallelogram  ADFE.  Hence  AF  represents  the 
resultant  of  the  three  forces  AB,  AC,  and  AD. 

Conversely,  the  force,  AF,  is  equivalent  to  the   three  ' 
components  AB,  AC,  and  AD. 

Let  P,  Q,  8  represent  the  three  forces  AB,  AC,  AD ;  i?, 
the  resultant;  «,  j3,  y,  the  angles  which  the  direction  of  R 
makes  with  the  directions  of  P,  Q,  S,  and  suppose  the 
forces  to  act  at  right  angles  with  each  other.     Then  since 

AF2  =  AB2  +  AC2  +  AD2, 
we  have  B*  =  P2  +  Q2  +  tf2; 

also,  cos  a  —  -jz, 

it 

Q 

cos  0  =  -g, 


a 

*x^ 


Fig.5 


(i) 


(2) 


cos  y  = 


R\ 


EXAMPLES.  29 

from  which  the  magnitude  and  direction  of  the  resultant 
are  determined. 

EXAMPLES. 

1.  Three  forces  of  5  lbs.,  3  lbs.,  and  2  lbs.,  respectively, 
act  upon  a  point  in  the  same  direction,  and  two  other  forces 
of  8  lbs.  and  9  lbs.  act  in  the  opposite  direction.  What 
single  force  will  keep  the  point  at  rest  ?  Ans.  7  lbs. 

2.  Two  forces  of  5^  lbs.  and  3|  lbs.,  applied  at  a  point, 
urge  it  in  one  direction ;  and.  a  force  of  2  lbs.,  applied  at 
the  same  point,  urges  it  in  the  opposite  direction.  What 
additional  force  is  necessary  to  preserve  equilibrium  ? 

Ans.  7  lbs. 

3.  If  a  force  of  13  lbs.  be  represented  by  a  line  of  6| 
inches,  what  line  will  represent  a  force  of  7 J  lbs.? 

Ans.  3f  inches. 

4.  Two  forces  whose  magnitudes  are  as  3  to  4,  acting  on 
a  point  at  right  angles  to  each  other,  produce  a  resultant  of 
20  lbs.;  required  the  component  forces. 

Ans.  12  lbs.  and  16  lbs. 

5.  Let  ABO  be  a  triangle,  and  D  the  middle  point  of 
the  side  BC.  If  the  three  forces  represented  in  magnitude 
and  direction  by  AB,  AC,  and  AD,  act  upon  the  point  A; 
find  the  direction  and  magnitude  of  the  resultant. 

Ans.  The  direction  is  in  the  line  AD,  and  the  magni- 
tude is  represented  by  3AD. 

6.  When  P  =  Q  and  0  =  60°,  find  R. 

Ans.  R  =  P  \/3. 

7.  When  P  =  Q  and  0  =  135°,  find  R. 


Ans.  R  =  P  V2  —  <v/& 

8.   When  P  =  Q  and  0  =  120°,  find  R. 

Am    R  =  P. 


30  RESOLUTION    OF   FORCES, 

9.  If  P  =  Q,  show  that  their  resultant  R  =  2P  cos  -• 

<i 

10.  If  P  =  8,  and  Q  =  10,  and  6  =  60°,  find  E. 

Ans.  R  =  2  a/61. 

11.  If  P  =  144,  72  =  145,  and  6  =  90°,  find  (J. 

Ans.    Q  =  17. 

12.  Two  forces  of  4  lbs.  and  3  V%  lbs.  act  at  an  angle  of 
45°,  and  a  third  force  of  \/42  lbs.  acts  at  right  angles  to 
their  plane  at  the  same  point ;  find  their  resultant. 

Ans.  10  lbs. 

35.  Resolution  of  Forces. — By  the  resolution  of  forces 
is  meant  the  process  of  finding  the  components  of  given  forces. 
We  have  seen  (Art.  30)  that  two  concurring  forces,  P  and 
Q  —  AB  and  AC,  (Fig.  2)  are  equivalent  to  a  single  force 
R  =  AD;  it  is  evident  then  that  the  single  force,  R,_ acting 
along  AD,  can  be  replaced  by  the  two  forces,  £  and  Q, 
represented  in  magnitude  and  direction  by  two  adjacent 
sides  of  a  parallelogram,  of  which  AD  is  the  diagonal. 

Since  an  infinite  number  of  parallelograms,  of  each  of 
which  AD  is  the  diagonal,  can  be  constructed,  it  follows 
that  a  single  force,  R,  can  be  resolved  into  two  other  forces 
in  an  infinite  number  of  ways. 

Also,  each  of  the  forces  AB,  AC,  may  be  resolved  into 
two  others,  in  a  way  similar  to  that  by  which  AD  was 
resolved  into  two ;  and  so  on  to  any  extent.  Hence,  a  single 
force  may  be  resolved  into  any  number  of  forces,  whose 
combined  action  is  equivalent  to  the  original  force. 

v  p 

Cor. — The  most  convenient  compo- 
nents into  which  a  force  can  be  resolved 
are  those  whose  directions  are  at  right 
angles  to  each  other.  Thus,  let  OX 
and  OY  be  any  two  lines  at  right 
angles  to  each  other,  and  P  any  force  acting  at  0  in  the 


MAGNITUDE    AND    DIRECTION    OF    RESULTANT. 


31 


plane  XOY.  Then  completing  the  rectangle  OMPN  w< 
find  the  components  of  P  along  the  axes  OX  and  0  Y  to  be 
0Jf  and  ON,  which  denote  by  X  and  Y.  Then  we  have 
clearly 

X=Pcos«,  )«  . 

Y  =  P  sin  «;  f  K  } 

where  «  is  the  angle  which  the  direction  of  P  makes  with 
OX.  These  components  X  and  Y  are  called  the  rect-' 
angular  components.  The  rectangular  component  of  a 
force,  P,  along  a  right  line  is  Px  cosine  of  angle  between 
line  and  direction  of  P. 

In  strictness,  when  we  speak  of  the  component  of  a  given 
force  along  a  certain  line,  it  is  necessary  to  mention  the 
other  line  along  which  the  other  component  acts.  In  this 
work,  unless  otherwise  expressed,  the  component  of  a  force 
along  any  line  will  be  understood  to  be  its  rectangular 
component ;  i.  e.,  the  resolution  will  be  made  along  this  line 
and  the  line  perpendicular  to  it. 

36.  To  find  the  Magnitude  and  Direction  of  the 
Resultant  of  any  number  of  Concurring  Forces  in 
one  Plane. — When  there  are  several  concurring  forces,  the 
condition  of  their  equilibrium  may  be  expressed  as  in 
Art.  33,  Cors.  1  and  2.  But  in  practice  we  obtain  much 
simpler  results  by  using  the  principle  of  the  Resolution  of 
Forces  (Art.  35),  than  those  given  by  the  principle  of 
Composition  of  Forces. 

Let  0  be  the  point  at  which  all 
the  forces  act.  Through  0  draw  the 
rectangular  axes  XX',  YY'.  Let 
P1,  P2,  P3,  etc.,  be  the  forces  and 
«15  (c2,  ccs,  etc.,  be  the  angles  which 
their  directions  make  with  the  axis 
of  x.  , 

Now  resolve  each  force  into  its  two  Fia-7 

components  along  the  axes  of  x  and   y.     Then   the  com- 


32        MAGNITUDE    AND    DIRECTION    OF   RESULTANT. 

ponents  along  the  axis  of  x  (^-components)  are  (Art. 
35,  Cor.),  Px  cos  cct,  P2  cos  «2,  P3  cos  «3,  etc.,  and  those 
along  the  axis  of  y  are  Pt  sin  at,  P2  sin  a2,  P3  sin  «3, 
etc. ;  and  therefore  if  X  and  F  denote  the  algebraic  sum  of 
the  ^-components  and  ^/-components  respectively,  we  have 

X  =  Pt  cos  at  +  P2  cos  cc2  -f  P3  cos  a3  -f  etc.  I       , 
=  SP  cos  «,  S      (  ' 

Y=  Pt  sin  «!  +  P2  sin  «2  +  P3  sin  a3  +  etc.  )       .  . 
=  2P  sin  «.  f      W 

Let  P  be  the  resultant  of  all  the  forces  acting  at  0,  and  8 
the  angle  which  it  makes  with  the  axis  of  x ;  then  resolving 
P  into  its  x-  and  ^/-components,  we  have 


P  cos  0  —  X  =  2P  cos 
P  sin  6  =  Y  =  2P  sin 


:(       <»> 


•.     P2  =  X2  +  F2;  tan  0  =  ^,  (4) 

^± 

which    determines  the  magnitude   and  direction   of    the 
resultant. 

Sch. — Regarding  OX  and  OF  as  positive  and  OX1  and 
OY1  as  negative  as  in  Anal.  Geom.,  we  see  that  Oxu  Oyt, 
Oy2  are  positive,  and  Ox2,  Ox3,  0y3  are  negative.  The 
forces  may  always  be  considered  as  positive,  and  hence  the 
signs  of  the  components  in  (1)  and  (2)  will  be  the  same  as 
those  of  the  trigonometric  functions.  Thus,  since  a2  is 
>  90°  and  <  180°  its  sine  is  positive  and  cosine  is  negative; 
since  cc3  is  >  180°  and  <  270°  both  its  sine  and  cosine  are 
negative. 

37.  The  Conditions  of  Equilibrium  for  any  number 
of  Concurring  Forces  in  one  Plane.— For  the  equilibrium 
of  the  forces  we  must  have  P  =  0.  Hence  (4)  of  Art.  36 
becomes 

X*  +  F2  =  0.  (1J 


EXAMPLES.  33 

Now  (1)  cannot  be  satisfied  so  long  as  X  and  Y  are  real 
quantities  unless  X  =  0,  Y  =  0 ;  therefore, 

X  =  ZP  cos  a  =  0  and  F  =  £P  sin  «  =  0.        (2) 

Hence  these  are  the  two  necessary  and  sufficient  conditions 
for  the  equilibrium  of  the  forces ;  that  is,  the  algebraic  sum 
of  the  rectangular  components  of  the  forces,  along  each  of 
tivo  right  lines  at  right  angles  to  each  other,  in  the  plane  of 
the  forces,  is  equal  to  zero.  As  the  conditions  of  equilibrium 
must  be  independent  of  the  S}7stem  of  co-ordinate  axes,  it 
follows  that,  if  auy  number  of  concurring  forces  in  one 
plane  are  in  equilibrium,  the  algebraic  sum  of  the  rectan- 
gular components  of  the  forces  along  every  right  line  in  their 
plane  is  zero. 

EXAMPLES. 

1.  Given  four  equal  concurring  forces  whose  directions 
are  inclined  to  the  axis  of  x  at  angles  of  15°,  75°,  135°, 
and  225°  ;  determine  the  magnitude  and  direction  of  their 
resultant. 

Let  each  force  be  equal  to  P ;  then 

X  =  P  cos  15°  +  P  cos  75°  +  P  cos  135°  +  P  cos  225° 

-  p3i~3 
S*     " 

Y  =  P  sin  15°  +  P  sin  75°  +  P  sin  135°  +  P  sin  225° 

=  P(t>*. 

.•.    E  =  P(5-2A/3)i# 

3* 

tan  0  =  -= • 

3^-2 

2.  Given  two  equal  concurring  forces,  P,  whose  direc- 
tions are  inclined  to  the  axis  of  x  at  angles  of  30°  and  315°; 
find  their  resultant.  Ans.  R  =  1.59  P. 

2* 


34  coNCURnma  forces. 

3.  Given  three  concurring  forces  of  4,  5,  and  6  lbs., 
whose  directions  are  inclined  to  the  axis  of  x  at  angles  oi 
0°,  60°,  and  135°  respectively;  find  their  resultant. 

Ans.  R  =  ^97  +  15  V6  —  39  \7I. 

4.  Given  three  equal  concurring  forces,  P,  whose  direc- 
tions are  inclined  to  the  axis  of  x  at  angles  of  30°,  60°,  and 
165°  ;  find  their  resultant.  Ans.  R  =  1.67  P. 

5.  Given  three  concurring  forces,  100,  50,  and  200  lbs., 
whose  directions  are  inclined  to  the  axis  of  x  at  angles  of 
0°,  60°,  and  180° ;  find  the  magnitude  and  direction  of 
their  resultant.  Ans.  R  =  86.6  lbs.;  0  =  150°. 

38.  To  find  the  Magnitude  and  Direction  of  the 
Resultant  of  any  number  of  Concurring  Forces  in 
Space. — Let  Pt,  P2,  P3,  etc.,  be  the  forces,  and  the 
whole  be  referred  to  a  system  of  rectangular  co-ordinates. 
Let  alf  (3t,  ylf  be  the  angles  which  the  direction  of  Px 
makes  with  three  rectangular  axes  drawn  through  the  point 
of  application  ;  let  «2,  j38,  y2,  be  the  angles  which  the  direc- 
tion of  Pg  makes  with  the  same  axes ;  «3,  j33,  y3,  the 
angles  which  P3  makes  with  the  same  axes,  etc.  Resolve 
these  forces  along  the  co-ordinate  axes  (Art.  35)  ;  the  com*- 
ponents  of  Pd  along  the  axes  are  Pt  cos  alf  Px  cos  j3,,  Px 
cos  y2.  Resolve  each  of  the  other  forces  in  the  same  way, 
and  let  X,  Y,  Z,  be  the  algebraic  sums  of  the  components 
of  the  forces  along  the  axes  of  x,  y,  and  z,  respectively ; 
then  we  have 

X  =  Pj  cos  ax  -f  P2  C0S  «2  +  -P&  C0S  <*3  +  e^c« 
=  XP  cos  a. 

Y  =  Pj  cos  0!  +  P2  cos  (32  4-  P3  cos  j33  +  etc.. 

=  SP  cos  j3.  (  ( 

Z  =  Pj  cos  yt  +  P2  cos  y8  +  P8  cos  y3  -f-  etc. 

=  £P  cos  y. 


CONDITIONS    OF  EQUILIBRIUM.  35 

Let  R  be  the  resultant  of  all  the  forces;  and  let  the 
angles  which  its  direction  makes  with  the  three  axes  be  a, 
b,  c ;  then  as  the  resolved  parts  of  R  along  the  three  co-or- 
dinate axes  are  equal  to  the  §tim  of  the  resolved  parts  of 
the  several  components  along  the  same  axes,  we  have 

R  cos  a  =  X,     R  cos  b  =  Y9     R  cos  c  =  Z.       (2j 

Squaring,  and  adding,  we  get 

R>  =  X2  +  F2  +  Z*;  (3) 

COS  a  =  -75,      COS  O  =  -75,      COS  C  =  -73;  (4) 

K  K  K 

which  determines  the  magnitude  of  the  resultant  of  any 
system  of  forces  in  space  and  the  angles  its  direction  makes 
with  three  rectangular  axes. 

39.  The  Conditions  of  Equilibrium  for  any  num- 
ber of  Concurring  Forces  in  Space. — If  the  forces  are 
in  equilibrium,  R  =  0  ;  therefore  (3)  of  Art.  38  becomes 

jT2  +  r*  +  z2  =  0. 

But  as  every  square  is  essentially  positive,  this  cannot  be 
unless  X  =  0,  Y  =  0,  Z  =  0  ;  and  therefore 

2P  cos  a  =  0,     EP  cos  0  =  0,     IP  cos  y  =  0 ;     (1) 

and  these  are  the  conditions  among  the  forces  that  they 
may  be  in  equilibrium  ;  that  is,  the  sum  of  the  components 
of  the  forces  along  each  of  the  three  co-ordinate  axes  is 
equal  to  zero. 

40.  Tension  of  a  String. — By  the  tension  of  a  string 
is  meant  the  pull  along  its  fibres  which,  at  any  point,  tends 
to  stretch  or  break  the  string.  In  the  application  of  the 
preceding  principles  the  string  or  cord  is  often  used  as  a 


36  EXAMPLES. 

means  of  communicating  force.  A  string  is  said  to  be  per- 
fectly flexible  when  any  force,  however  small,  which  is 
applied  otherwise  than  along  the  direction  of  the  string, 
will  change  its  form.  In  this  work  the  string  will  be 
regarded  as  perfectly  flexible,  inextensible,  and  without 
weight. 

If  such  a  string  be  kept  in  equilibrium  by  twro  forces, 
one  at  each  end,  it  is  clear  that  these  forces  must  be  equal 
and  act  in  opposite  directions,  so  that  the  string  assumes 
the  form  of  a  straight  line  in  the  direction  of  the  forces. 
In  this  case  the  tension  of  the  string  is  the  same  through- 
out, and  is  measured  by  the  force  applied  at  one  end  ;  and 
if  it  passes  over  a  smooth  peg,  or  over  any  number  of 
smooth  surfaces,  its  tension  is  the  same  at  all  of  its  points. 
If  the  string  should  be  knotted  at  any  of  its  points  to  other 
strings,  we  must  regard  its  continuity  as  broken,  and  the 
tension,  in  this  case,  will  not  be  the  same  in  the  two  por- 
tions which  start  from  the  knot. 

EXAM  PLES. 

1.  A  and  B  (Fig.  8)  are  two  fixed 
points  in  a  horizontal  line ;  at  A  is 
fastened  a  string  of  length  b,  with  a 
smooth  ring  at  its  other  extremity,  0, 
through  which  passes  anotherstring  with 
one  end  fastened  at  B,  the  other  end  of  '9' 

which  is  attached  to  a  given  weight  W  ;  it  is  required  to 
determine  the  position  of  C. 

Before  setting  about  the  solution  of  statical  problems  of 
this  kind,  the  student  will  clear  the  ground  before  him,  and 
greatly  simplify  his  labor  by  asking  himself  the  following 
questions  :  (1)  What  lines  are  there  in  the  figure  whose 
lengths  are  already  given  ?  (2)  What  forces  are  there 
whose  magnitudes  are  already  given,  and  what  are  the 
forces   whose  magnitudes  are  yet  unknown?     (3)  What 


EXAMPLES.  3? 

variable  lines  or  angles  in  the  figure  would,  if  they  were 
known,  determine  the  required  position  of  0  ? 

Now  in  this  problem,  (1)  the  linear  magnitudes  which 
are  given  are  the  lines  AB  aud  AC.  (2)  The  forces  acting 
at  the  point  0  to  keep  it  at  rest  are  the  weight  W,  a  ten- 
sion in  the  string  CB,  and  another  tension  in  the  string 

CA.  Of  these    W   is  given,    and   so   is   the  tension   in 

CB,  which  must  also  be  equal  to  W,  since  the  ring  is 
smooth  and  the  tension  therefore  of  WCB  is  the  same 
throughout  and  of  course  equal  to  W.  But  as  yet  there  is 
nothing  determined  about  the  magnitude  of  the  tension  in 
C A.  And  (3)  the  angle  of  inclination  of  the  string  CA  to 
the  horizon  would,  if  known,  at  once  determine  the  posi- 
tion of  C.  For  if  this  angle  is  known,  we  can  draw  AC  of 
the  given  length ;  then  joining  C  to  B,  the  position  of  the 
system  is  completely  known. 

Let  AB  =  a,  AC  =  b,  CAB  =  0,  CBA  =  0,  and  the 
tension  of  the  string  AC  =  T.  Then,  for  the  equilibrium 
of  the  point  C  under  the  action  of  the  three  forces,  W,  W, 
and  T,  we  apply  (2)  of  Art.  37,  and  resolve  the  forces 
horizontally  and  vertically;  and  equate  those  acting  towards 
the  right-hand  to  those  acting  towards  the  left ;  and  those 
acting  upwards  to  those  acting  downwards.  Then  the 
horizontal  and  vertical  forces  are  respectively 

IF  cos  0  =  Tcosd; 

Wsm<p  +  I7  sin  0  =  W. 

Eliminating  T  we  have 

cos  0  =  sin  (0  -f-  0)  ; 

.-.     2(9  +  0  =  90°.  (1) 

Also,  from  trigonometry  we  have 


sin  (0  +  0)  _  a 
sin  0         ~  b 


(2) 


38  EXAMPLES. 

from  (1)  and  (2)  6  and  0  may  be  found ;  and  therefore  T 
may  be  found ;  and  thus  all  the  circumstances  of  the 
problem  are  determined. 

2.  One  end  of  a  string  is  attached  to 
a  fixed  point,  A,  (Fig.  9) ;  the  string,  after 
passing  over  a  smooth  peg,  B,  sustains  a 
given  weight,  P,  at  its  other  extremity, 
and  to  a  given  poiut;  C,  in  the  string  is 
knotted  a  given  weight,  W.  Find  the  posi- 
tion of  equilibrium. 

The  entire  length  of  the  string,  AOBP,  is  of  no  conse- 
quence, since  it  is  clear  that,  once  equilibrium  is  estab- 
lished, P  might  be  suspended  from  a  point  at  any  distance 
whatever  from  B.  The  forces  acting  at  the  point,  C,  are 
the  given  weight,  W,  the  tension  in  the  string,  CB,  which, 
since  the  peg  is  smooth,  is  P,  and  the  tension  in  the  string 
OA,  which  is  unknown. 

Let  AB  =  a,  AC  =  b,  CAB  =  6,  CBA  =  0,  and  the 
tension  of  the  string,  AC  =  T.  Then  for  the  equilibrium 
of  the  point  C,  we  have  (Art.  32), 


p  cos  e 


(i) 


W       sin  (0  +  0) ' 
also,  from  the  geometry  of  the  figure,  we  have 

b  sin  (6  -f  0)  =  a  sin  0.  (2) 

From  (1)  and  (2)  we  get 

P  _  b  cos  6 

W  ~  a  sin  0 ' 

or  sin  0  =  — =:  cos  8 ; 

aP^ 

V a* PK—  b2W*  cos2  6 


.        Va*P\- 


EXAMPLES.  39 

Expanding  sin  (0  -f  <p)  in  {'I),  and  substituting  in  it  these 
values  of  sin  </>  and  cos  <f),  and  reducing,  we  have  the 
equation 

cos3 '  -  -     Ta^rt  -  ™s2 «  +  HF«  =  °' 

from  which  0  may  be  found.  (See  Minchin's  Statics, 
p.  29.) 

3.  If,  in  the  last  example,  the  weight,  W,  instead  of 
being  knotted  to  the  string  at  C,  is  suspended  from  a 
smooth  ring  which  is  at  liberty  to  slide  along,  the  string, 
ACB,  find  the  position  of  equilibrium. 

W 
Ans.  sin  6  =  — =. 
ZJr 

41.  Equilibrium  of  Concurring  Forces  on  a 
Smooth  Plane. — If  a  particle  be  kept  at  rest  on  a  smooth 
surface,  plane  or  curved,  by  the  action  of  any  number  of 
forces  applied  to  it,  the  resultant  of  these  forces  must  be  in 
the  direction  of  the  normal  to  the  surface  at  the  point 
where  the  particle  is  situated,  and  must  be  equivalent  to 
the  pressure  which  the  surface  sustains.  For,  if  the 
resultant  had  any  other  direction  it  could  be  resolved  into 
two  components,  one  in  the  direction  of  the  normal  and  the 
other  in  the  direction  of  a  tangent ;  the  first  of  these  would 
be  opposed  by  the  reaction  of  the  surface ;  the  second  being 
unopposed,  would  cause  the  particle  to  move.  Hence,  we 
may  dispense  with  the  plane  altogether,  and  regard  its 
normal  reaction  as  one  of  the  forces  by  which  the  particle 
is  kept  at  rest.  Therefore  if  the  particle  on  which  the 
statical  forces  act  be  on  a  smooth  plane  surface,  the  case  is 
the  same  as  that  treated  in  Art.  39,  yiz.,  equilibrium  of  a 
particle  acted  upon  by  any  number  of  forces ;  and  in  writ- 
ing down  the  equations  of  equilibrium,  we  merely  have  to 
include  the  normal  reaction  of  the  plane  among  all  the 
others. 


40 


EXAMPLES. 


EXAMPLES. 

1.  A  heavy  particle  is  placed  on  a 
smooth  inclined  plane,  AB,  (Fig.  10), 
and  is  sustained  by  a  force,  P,  which 
acts  along  AB  in  the  vertical  plane 
which  is  at  right  angles  to  AB ;  find 
P,  and  also  the  pressure  on  the  in- 
clined plane. 

The  only  effect  of  the  inclined  plane  is  to  produce  a 
normal  reaction,  R,  on  the  particle.  Hence  if  we  intro- 
duce this  force,  we  may  imagine  the  plane  removed. 

Let  W  be  the  weight  of  the  particle,  and  a  the  inclina- 
tion of  the  plane  to  the  horizon. 

Resolving  the  forces  along,  and  perpendicular  to  AB, 
since  the  lines  along  which  forces  may  be  resolved  are 
arbitrary  (Art.  37),  we  have  successively, 


P  —  W  sin  a  =  0,     or     P  =  W  sin  a  ; 


and 


R  —  W  cos  a  =  0,     or     R  =  W  cos  a. 


If,  for  example,  the  weight  of  the  particle  is  4  oz.,  and 
the  inclination  of  the  plane  30°,  there  will  be  a  normal 
pressure  of  2 a/3  oz.  on  the  plane,  and  the  force,  P,  will 
be  2  oz. 

2.  In  the  previous  example,  if  P  act  horizontally,  find 
its  magnitude,  and  also  that  of  R. 

Resolving  along  AB  and  perpendicular  to  it,  we  have 
successively, 


P  cos  a  —  W  sin  a  =  0,     or     P  =  W  tan  a  ; 

W 


and      P  sin  a  +  W  cos  a  —  R  =  0,     .*.     R  = 


cos  a 


CONDITIONS   OF  EQUILIBRIUM.  41 

3.  If  the  particle  is  sustained  by  a  force,  P,  making  a 
given  angle,  0,  with  the  inclined  plane,  find  the  magnitude 
of  this  force,  and  of  the  pressure  on  the  plane,  all  the  forces 
acting  in  the  same  vertical  plane. 

Eesolving  along  and  perpendicular  to  the  plane  succes- 
sively, we  have 

P  cos  0  —  W  sin  a  =  0, 
and  R  +  P  sin  d  —  W  cos  a  =  0, 

from  which  we  obtain 

P=W——  •     R=  WC08(a  +  e) 
cos  0  '  cos  6 

Rem. — The  advantage  of  a  judicious  selection  of  direc- 
tions for  the  resolution  of  the  forces  is  evident.  By  resolv- 
ing at  right  angles  to  one  of  the  unknown  forces,  we 
obtain  an  equation  free  from  that  force ;  whereas  if  the 
directions  are  selected  at  random,  all  of  the  forces  will 
enter  each  equation,  which  will  make  the  solution  less 
simple. 

The  student  will  observe  that  these  values  of  P  and  R 
could  have  been  obtained  at  once,  without  resolution,  by 
Art.  32. 

42.  Conditions  of  Equilibrium  for  any  number  of 
Concurring  Forces  when  the  particle  on  which  they 
act  is  Constrained  to  Remain  on  a  Given  Smooth 
Surface. — If  a  particle  be  kept  at  rest  on  a  smooth  sur- 
face by  the  action  of  any  number  of  forces  applied  to  it, 
the  resultant  of  these  forces  must  be  in  the  direction  of  the 
normal  to  the  surface  at  the  point  where  the  particle  is 
situated,  and  must  be  equivalent  to  the  pressure  which  the 
surface  sustains  (Art.  41).  Hence  since  the  resultant  is  in 
the  direction  of  the  normal,  and  is  destroyed  by  the  reac- 


42  CONDITIONS   OF  EQUILIBRIUM. 

tion  of  the  surface,  we  may  regard   this  reaction   as   an 
additional  force  directly  opposed  to  the  normal  force. 

Let  N  be  the  normal  reaction  of  the  surface,  and  «,  j3,  y, 
the  angles  which  N  makes  with  the  co-ordinate  axes  of  x, 
y,  and  z,  respectively.  Let  X,  Y,  Z,  be  the  sum  of  the 
components  of  all  the  other  forces  resolved  parallel  to  the 
three  axes  respectively.  The  reaction  N  may  be  considered 
a  new  force,  which,  with  the  other  forces,  keeps  the  parti- 
cle in  equilibrium.  Therefore,  resolving  N  parallel  to  the 
three  axes,  we  have  (Art.  39), 

X  +  N  cos  a  =  0,  ) 

Y  +  AT  cos  [3  =  0,  V  (1) 

Z  +  N  cos  y  =  0.  ) 

Let  u  =zf(x,  y,  z)  =  0,  be  the  equation  of  the  given 
surface,  and  x,  y,  z  the  co-ordinates  of  the  particle  to 
which  the  forces  are  applied.  We  have  (Anal.  Geom., 
Art.  175), 

a' 


cos  a 


VV  2  +   1/2  +   1  * 


cos  j3  =  — ==A  (2) 


i 

cos  y  = 


Vet'*  +  b*  +  l?y 

where  a'  and  V  are  the  tangents  of  the  angles  which  the 
projections  of  the  normal,  N,  on  the  co-ordinate  planes  xz 
and  yz  make  with  the  axis  of  z.  Since  the  normal  is  per- 
pendicular to  the  plane  tangent  to  the  surface  at  (x,  y,  z), 
the  projections  of  the  normal  are  perpendicular  to  the 
traces  of  the  plane.     Therefore   (Anal.  Geom.,    Art.    27, 

(3) 
(i) 


Cor.  1),  we  have 

1  +  aa!  =  0, 

and 

1  +  W  =  0 ; 

CONDITIONS   OF  EQUILIBRIUM.  43 

in  which 

dx        ,       dx'      7        dii      ,,       dv' 
dz  dz  dz  dz 

(Calculus,  Art.  56^.)     Substituting  in  (3)  and  (4),  we  have 


dx     dx' 

1  +  Tz'  ^-U; 

and 

from  which 

dx' 

.       dz'  ~~ 

du 

dz         dx   lf   ,    . 
—  -=-  =  — —  (Cal.  Art. 
dx        du   v 

dz 

dv, 

(5) 


and  %  =  -d*=*f  =  V.  (6) 

dz  dy        du 

dz 

Substituting  these  values  of  a'  and  V  in  (2)  and  multiply- 

du 
ing  both  terms  of  the  fraction  by  -?-,  we  have 


cos  a  = 


du 

dx 


/(du\*      iduV      /du\2' 

v  vs) +  y + u 


cos  (3  =  — ■  ^  —  -         g,  \  (7) 

//du\*     (du\*      /du\*' ' 

v  y + y + y 


cos  y  = 


/  (du\2      {du\2      (du\i9 

v  «) + y + W)  , 


44 


CONDITIONS    OF   EQUILIBRIUM, 


which  give  the  values  of  the  direction  cosines  of  the  normal 
at  (x,  y,  z). 

Putting  the  denominator  equal  to  Q,  for  shortness,  and 
substituting  in  (1)  and  transposing,  we  have 


y           N    du 
JL"  ~~Q  *  di' 

(8) 

v           N    du 

(9) 

7           N    du 

~Q  '  dz 

(10) 

X 

du 

Y 

du 

Z 

~  du9 

dx 

dy 

dz 

Eliminating  N  between  these  three  equations,  we  obtain 
the  two  independent  equations, 


(ii) 


which  express  the  conditions  that  must  exist  among  the 
applied  forces  and  their  directions  in  order  that  their 
resultant  may  be  normal  to  the  surface,  i.  e.,  that  there  may 
be  equilibrium.  If  these  two  equations  are  not  satisfied, 
equilibrium  on  the  surface  cannot  exist.  Hence  the  point 
on  a  given  surface,  at  which  a  given  particle  under  the 
action  of  given  forces  will  rest  in  equilibrium,  is  the  point 
at  which  equations  (11)  are  satisfied. 


Cor.  1. — Squaring  equations  (8),  (9),  (10)  and  adding,  we 

du\*' 


get 

X2  +  Y*  +  Z*  =  N2 


(du\*       (du\2       (duV 
\dx)        \dyl        \dz) 

r\o       ~T"        r\o         i  na 


L  Q2 


«■  J 


=  jt*; 


JV  =  VX1  +  F2  +  Z\ 


(12) 


EXAMPLES.  45 

which  is  the  value  of  the  normal  resistance  of  the  surface 
and  is  precisely  the  same  as  the  resultant  of  the  acting 
forces,  as  it  clearly  should  be  ;  but  this  resistance  must  act 
in  the  direction  opposite  to  that  of  the  resultant. 

Cor.  2. — Multiplying  (8),  (9),  (10)  by  dx,  dy,  dz,  respec- 
tively, and  adding,  and  remembering  that  the  total  differ- 
ential of  u  =  0  is  zero,  we  get 

Xdx  +  Ydy  +  Zdz  =  0,  (13) 

which  is  an  equation  of  condition  for  equilibrium.  If  (13) 
cannot  be  satisfied  at  any  point  of  the  surface,  equilibrium 
is  impossible. 

Cor.  3. — If  the  forces  all  act  in  one  plane,  the  surface 
becomes  a  plane  curve ;  let  this  curve  be  in  the  plane  xy, 
then  z  =  0;  therefore  (11)  and  (13)  become 

~d^  =  ~d^f  (U) 

dx         dy 

and  Xdx  +  Ydy  =  0,  (15) 

in  which  (14)  or  (15)  may  be  used  according  as  the  equation 
of  the  curve  is  given  as  an  implicit  or  explicit  function. 

EXAMPLES. 

1.  A  particle  is  placed  on  the  surface  of  an  ellipsoid,  and 
is  acted  on  by  attracting  forces  which  vary  directly  as  the 
distance  of  the  particle  from  the  principal  planes*  of  sec- 
tion ;  it  is  required  to  determine  the  position  of  equilibrium. 

Let  the  equation  of  the  ellipsoid  be 

r3         ?/2         *2 

u=f(x,y,z)  =  ^  +  1+^-1=0; 

*  Planes  of  xy,  yz,  zx. 


46  EXAMPLES. 

du  _  2x      du  _  2y      du  _  2z 
'*•     dx~~W     dy~~~l?7     dzZ~J; 

.a,nd  let   the   sc-,   y-,    and   2-coinponents    of   the  forces   be 
respectively, 

X  =  —  uxx9    Y  =  —  u2y,  Z  =  —  u3z  ; 

then  (11)  will  give 

uxa2  =  u2tP'=  u3c2; 
which  may  be  put  in  the  form 

ux        u2        u3         u1  +  u2  -f  u3 


a-2  1-2  c~2  ^-2  +  J-2  +  c- 


If  these  conditions  are  fulfilled,  the  particle  will  rest  at  all 
points  of  the  surface. 

2.  Again,  take  the  same  surface,  and  let  the  forces  vary 
inversely  as  the  distances  of  the  point  from  the  principal 
planes;  it  is  required  to  determine  the  position  of  equili- 
brium. 

Here     X=-^,     F=  —  5t,     Z=-^; 
x  y  z 

therefore  (11)  becomes 

z2        f        z* 

a?  __&_  __  ?_  __  1  _  1 

ut        u2        u3       ux  +  u2  -j-  u3       u 

by  putting  u  for  ux  +  u2  +  u3, 


EXAMPLES.  47 

which  in  (12)  gives 

X1  if  Z2 


3.  A  particle  is  placed  inside  a  smooth  sphere  on  the  con- 
cave surface,  and  is  acted  on  by  gravity  and  by  a  repulsive 
force  which  varies  inversely  as  the  square  of  the  distance 
from  the  lowest  point  of  the  sphere ;  find  the  position  of 
equilibrium  of  the  particle. 

Let  the  lowest  point  of  the  sphere  be  taken  for  the  origin 
of  co-ordinates,  and  let  the  axis  of  z  be  vertical,  and  posi- 
tive upwards;  then  the  equation  of  the  sphere,  whose 
radius  is  a,  is 

x2  +  y%  +  Z2  _  2az  =  0. 

Let  W  =  the  weight  of  the  particle,  and  r  =  the  distance 
of  it  from  the  lowest  point ;  then 


/y-2 


X2  +  y2  +  Z2  —  2az. 

Also,  let  the  repulsive  force  at  the  unit's  distance  =  u  \ 
then  at  the  distance  r  it  will  be 


u 

u 
~  2az> 

TT 

u 

X 

X  = 

2az 

r9 

Y  = 

u 

2az 

y 

r9 

Z  = 

71 

2az   ' 

z 
r 

w. 


48 


EXAMPLES. 


Let  N  =  the  normal  pressure  of  the  curve  ;  then  (8)  and 
(10)  give 


U         X  ltX 

—->  -  +  N- 
2az     r  a 


0, 


2az     r  a 


from  which  we  have 


r3  = 


ua 
W 


,t 


2a»WV 


whence  the  position  of  the  particle  is  known  for  a  given 
weight,  and  for  a  given  value  of  u,  (See  Price's  Anal. 
Mechanics,  Vol.  I,  p.  39.) 

4.  Two  weights,  P  and  Q,  are  fastened  to  the  ends  of  a 
string,  (Fig.  11),  which  passes  over  a  pulley,  0 ;  and  Q 
hangs  freely  when  P  rests  on  a  plane  curve,  AP,  in  a 
vertical  plane  ;  it  is  required  to  find  the  position  of  equili- 
brium when  the  curve  is  given. 

The  forces  which  act  on  P  are  (1)  the 
tension  of  the  string  in  the  line  OP,  which 
is  equal  to  the  weight  of  Q,  (2)  the  weight 
of  P  acting  vertically  downwards,  (3)  the 
normal  reaction  of  the  curve  R. 

Let  0  be  the  origin  of  co-ordinates,  and 
the  axis  of  x  vertical  and  positive  down- 
wards. Let  OM  =  x,  MP  =  y,  OP  =  r, 
POM  =6,  OA  =  a.     Then, 

X=  P  -  Qcosd-E^, 


Fig.ll 


Q  sin  0  +  R 


dx  # 
ds' 


EXAMPLES.  49 

therefore  from  (15)  we  have 

(P  —  Q  cos  0)  dx  —  Q  sin  Q  dy  =  0, 

_  ,  ^  x  dx  -f  y  dy 

or  Pdx  —  Q — '   3    u  =  0. 

r 

But  since  x2  +  y2  =  r2, 

we  have  xdx  -{-  y  dy  =  r  dr  ; 

.'.     Pdx—  Qdr  =  0;  (1) 

which  is  the  condition  that  must  be  satisfied  by  P,  Q,  and 
the  equation  of  the  curve. 

5.   Required  the  equation  of  -the  curve,  on  all  points  of 
which  P  will  rest. 

Integrating  (1)  of  Ex.  4,  we  have 

Px  -  Qr  =  0.  (1) 

But  since  P  is  to  rest  at  all  points  of  the  curve,  this  equa- 
tion must  be  satisfied  when  P  is  at  A,  from  which  we  get 
x  =  r  =  a ;  therefore  (1)  becomes 

Pa  —  Qa  =  C; 
which  in  (1)  gives 

*r  —  — • 

P  ' 

1  —  yr  cos  6 

which  is  the  equation  of  a  conic  section,  of  which  the  focus 
is  at  the  pole  0 ;  -and  is  an  ellipse,  parabola,  or  hyperbola, 
according  as  P  <,  =,  or  >   Q. 
3 


50  EXAMPLES, 


EXAMPLES. 

1.  Two  forces  of  10  and  20  lbs.  act  on  a  particle  at  an 
angle  of  60° ;  find  the  resultant.  Ans.  26.5  lbs. 

2.  The  resultant  of  two  forces  is  10  lbs. ;  one  of  the 
forces  is  8  lbs.,  and  the  other  is  inclined  to  the  resultant  at 
an  angle  of  36°.  Find  it,  and  also  find  the  angle  between 
the  two  forces.  (There  are  two  solutions,  this  being  the 
ambiguous  case  in  the  solution  of  a  triangle.) 

Ans.  Force  is  2.66  lbs.,  or  13.52  lbs.  Angle  is  47°  17' 
05",  or  132°  42'  55". 

3.  A  point  is  kept  at  rest  by  forces  of  6,  8,  11  lbs. 
Find  the  angle  between  the  forces  6  and  8. 

Ans.  77°  21' 52". 

4.  The  directions  of  two  forces  acting  at  a  point  are 
inclined  to  each  other  (1)  at  an  angle  of  60°,  (2)  at  an 
angle  of  120°,  and  the  respective  resultants  are  as 
V7  :  a/3  ;  compare  the  magnitude  of  the  forces. 

Ans.  2  :  1. 

5.  Three  posts  are  placed  in  the  ground  so  as  to  form  an 
equilateral  triangle,  and  an  elastic  string  is  stretched  round 
them,  the  tension  of  which  is  6  lbs. ;  find  the  pressure  on 
each  post.  jw^  6  ^/s. 

6.  The  angle  between  two  unknown  forces  is  37°,  and 
their  resultant  divides  this  angle  into  31°  and  6°  ;  find  the 
ratio  of  the  component  forces.  Ans.  4.927  :  1. 

7.  If   two  equal  rafters  support  a  weight,    W,  at  their 

upper  ends,  required  the  compression    on   each.      Let  the 

length    of   each   rafter   be   a,  and  the  horizontal  distance 

between  their  lower  ends  be  b.  .  a)V 

Ans.  — 

\/4«2  —  & 


EXAMPLES.  51 

8.  Three  forces  act  at  a  point,  and  include  angles  of 
90°  and  45°.  The  first  two  forces  are  each  equal  to  2P, 
and  the  resultant  of  them  all  is  a/10P ;  find  the  third 
force.  Am.  P  VS. 

9.  Find  the  magnitude,  R,  and  direction,  0,  of  the 
resultant  of  the  three  forces,  Pt  —  30  lbs.,  P2  =  70  lbs., 
P3  =  50  lbs.,  the  angle  included  between  Pt  and  P2 
being  56°,  and  between  P2  and  P3  104°.  (It  is  generally 
convenient  to  take  the  action  line  of  one  of  the  forces  for 
the  axis  of  x.) 

Let  the  axis  of  x  coincide  with  the  direction  of  Pt ;  then 
(Art.  36),  we  have 

X  =  22.16  ;     7  =  75.13  ;     R  =  78.33  ;    6  =  73°  34'. 

10.  Three  forces  of  10  lbs.  each  act  at  the  same  point ; 
the  second  makes  an  angle  of  30°  with  the  first,  and  the 
third  makes  an  angle  of  60°  with  the  second  ;  find  the 
magnitude  of  the  resultant.  Ans.  24  lbs.,  nearly. 

11.  If  three  forces  of  99,  100,  and  101  units  respectively, 
act  on  a  point  at  angles  of  120°;  find  the  magnitude  of 
their  resultant,  and  its  inclination  to  the  force  of  100. 

Ans.  V3;  90°. 

12.  A  block  of  800  lbs.  is  so  situated  that  it  receives 
from  the  water  a  pressure  of  400  lbs.  in  a  south  direction, 
and  a  pressure  from  the  wind  of  100  lbs.  in  a  westerly 
direction  ;  required  the  magnitude  of  the  resultant  pres- 
sure, and  its  direction  with  the  vertical. 

Ans.  900  lbs. ;  27°  16'. 

13.  A  weight  of  40  lbs.  is  supported  by  two  strings,  one 
of  which  makes  an  angle  of  30°  with  the  vertical,  the  other 
45° ;  find  the  tension  in  each  string. 

Ans.  20  ( V6  -  V2) ;  40  (V%  —  1). 


52  EXAMPLES, 

14.  Two  forces,  P  and  P',  acting  along  the  diagonals  of 
a  parallelogram,  keep  it  at  rest  in  such  a  position  that  one 
of  its  sides  is  horizontal;  show  that 

P  sec  a   =  P'  sec  «  =  W  cosec  («  +  «'), 

where  IF  is  the  weight  of  the  parallelogram,  and  a  and  a 
the  angles  between  the  diagonals  and  the  horizontal  side. 

15.  Two  persons  pull  a  heavy  weight  by  ropes  inclined 
to  the  horizon  at  angles  of  60°  and  30°  with  forces  of 
160  lbs.  and  200  lbs.  The  angle  between  the  two  vertical 
planes  of  the  ropes  is  30° ;  find  the  single  horizontal  force 
that  would  produce  the  same  effect.  Ans.  245.8  lbs. 

16.  In  order  to  raise  vertically  a  heavy  weight  by  means 
of  a  rope  passing  over  a  fixed  pulley,  three  workmen  pull  at 
the  end  of  the  rope  with  forces  of  40  lbs.,  50  lbs.,  and 
100  lbs. ;  the  directions  of  these  forces  being  inclined  to 
the  horizon  at  an  angle  of  60°.  What  is  the  magnitude  of 
the  resultant  force  which  tends  directly  to  raise  the  wreight  ? 

Ans.  164.54  lbs. 

17.  Three  persons  pull  a  heavy  weight  by  cords  inclined 
to  the  horizon  at  an  angle  of  60°,  with  forces  of  100,  120, 
and  140  lbs.  The  three  vertical  planes  of  the  cords  are 
inclined  to  each  other  at  angles  of  30°;  find  the  single 
horizontal  force  that  would  produce  the  same  effect. 

Ans.   10  Vi45  +  72  Vs  lbs. 

18.  Two  forces,  P  and  Q,  acting  respectively  parallel  to 
the  base  and  length  of  an  inclined  plane,  will  each  singly 
sustain  on  it  a  particle  of  weight,  W;  to  determine  the 
weight  of  W. 

Let  a  =  inclination  of  the  plane  to  the  horizon ;  then 
resolving  in  each  case  along  the  plane,  so  that  the  normal 
pressures  may  not  enter  into  the  equations  (See  Rem.,  Ex.  3, 
Art.  41),  we  have 


EXAMPLES.  53 

P  cos  «  =  W  sin  a  ;     §  =  )T  sin  « ; 

PQ 


w  = 


(pa  _  0>)i 


v 


19.  A  cord  whose  length  is  21,  is  fastened  at  A  and  i?,  in 
the  same  horizontal  line,  at  a  distance  from  each  other 
equal  to  %a  ;  and  a  smooth  ring  upon  the  cord  sustains  a 
weight  )!'j  find  the  tension  of  the  cord. 

An*.   T  Wl 


2  VF  -  a? 

20.  A  heavy  particle,  whose  weight  is  W,  is  sustained  on 
a  smooth  inclined  plane  by  three  forces  applied  to  it,  each 

W 

equal  to  —  :  one  acts  vertically  upward,  another  horizon- 
o 

talk,  and  the  third  along  the  plane ;   find  the  inclination, 

«,  of  the  plane.  .  a       1 

Arts,  tan  -  =  -« 

2         2 

21.  A  body  whose  weight  is  10  Ids.  is  supported  on  a 
smooth  inclined  plane  by  a  force  of  2  lbs.  acting  along  the 
plane,  and  a  horizontal  force  of  5  lbs.  Find  the  inclination 
of  the  plane.  Ans.  sin-1  f . 

y 

22.  A  body  is  sustained  on  a  smooth  inclined  plane  (in- 
clination a)  by  a  force,  P,  acting  along  the  plane,  and  a 
horizontal  force,  Q.  When  the  inclination  is  halved,  and 
the  forces,  P  and  Q,  each  halved,  the  body  is  still  observed 

to  rest ;  find  the  ratio  of  P  to  Q.  .        P  na 

Ans.  -pr  =  2  cos2  T- 
Q  4 

23.  Two  weights,  P  and  Q,  (Fig.  12),  rest 
on  a  smooth  double-inclined  plane,  and  are 
attached  to  the  extremities  of  a  string 
which  passes  over  a  smooth  peg,  0,  at  a 
point  vertically  over  the  intersection  of  the 
planes,  the  peg  and  the  weights  being  in  a 


Fig.  12 


54 


EXAMPLES. 


vertical  plane.     Find  the  position  of  equilibrium,  if  I  =  the 
length  of  the  string  and  h  —  CO. 

Ansv  The  position  of  equilibrium  is  given  by  the  equa- 
tions 

p  sin  a  _    n  sin  (3 
cos  6  ~ "  ^  cos  <f> '  & 

cos  «      cos  (3  _    I 
sin  6       sin  0  ""  h 

24.  Two  weights,  P  and  Q,  connected  by  a  string, 
length  I,  rest  on  the  convex  side  of  a  smooth  vertical 
circle,  radius  a.  Find  the  position  of  equilibrium,  and 
show  that  the  heavier  weight  will  be  higher  up  on  the 
circle  than  the  lighter,  the  radius  of  the  circle  drawn  to  P 
making  an  angle  6  with  the  vertical  diameter. 

Ans.  Psin  6  =  Q  sin  (-  —  #)• 

25.  Two  weights,  P  and  Q,  connected  directly  by  a 
string  of  given  length,  rest  on  the  convex  side  of  a  smooth 
vertical  circle,  the  string  forming  a  chord  of  the  circle  ; 
find  the  position  of  equilibrium. 

Ans.  If  2a  is  the  angle  subtended  at  the  centre  of  the 
circle  by  the  string,  the  inclination,  6,  of  the  string  to  thi 
vertical  is  given  by  the  equation 

P—0 
cot  6  =  p     J*  tan  a. 


26.  Two  weights,  P  and  Q,  (Fig.  13), 
rest  on  the  concave  side  of  a  parabola 
whose  axis  is  horizontal,  and  are  con- 
nected by  a  string,  length  I,  which 
passes  over  a  smooth  peg  at  the  focus,  F. 
Find  the  position  of  equilibrium. 

Ans.  Let   6  =  the   angle   which   FP 


Fig. 13         Q 


EXAMPLES.  55 

makes  with  the  axis,  and  4w  =  the  latus  rectum  of  the 

parabola,  then 

6  PVl-  2m 

cot  -  =  — ===• 
2        Vm  (P*  +  C2) 

*  27.  A  particle  is  placed  on  the  convex  side  of  a  smooth 
ellipse,  and  is  acted  upon  by  two  forces,  F  aud  F',  towards 
the  foci,  and  a  force,  F",  towards  the  centre.  Find  the 
position  of  equilibrium. 

Aiis.  r  =  — .  where  r  =  the  distance  of  the  par- 

Vl  -  n2 
tide  from  the  centre  of  the  ellipse  ;  b  =  semi-minor  axis, 
F—  F' 


and  n  = 


F" 


28.  Let  the  curve,  (Fig.  11),  be  a  circle  in  which  the 
origin  and  pulley  are  at  a  distance,  a,  above  the  centre  of 
the  circle  ;  to  determine  the  posi'tion  of  equilibrium. 

Q 

Ans.  r  =  -pa. 

29.  Let  the  curve,  (Fig.  11),  be  a  hyperbola  in  which  the 
origin  and  pulley  are  at  the  centre,  0,  the  transverse  axis 
being  vertical  ;  to  determine  the  position  of  equilibrium. 

A  iP 

Ans.  x  = f 

30.  A  particle,  P,  is  acted  upon  by  two  forces  towards 

two  fixed  points,  S  and  H,  these  forces  being  -^-p  and  jjp, 

respectively ;  ^prove  that  P  will  rest  at  all  points  inside  a 
smooth  tube  in  the  form  of  a  curve  whose  equation  is  SP. 
PH  =  k2,  h  being  a  constant. 

31.  Two  weights,  P  and  Q,  connected  by  a  string,  rest 
on  the  convex  side  of  a  smooth  cycloid.  Find  the  position 
of  equilibrium. 


56  EXAMPLES. 

Ans.  If  I  =  the  length  of  the  string,  and  a  =  radius  of 
generating  circle,  the  position  of  equilibrium  is  defined  by 
the  equation 

e         q        i 


sin 


2       P  +  Q     4a' 


where  0  is  the  angle  between  the  vertical  and  the  radius  to 
the  point  on  the  generating  circle  which  corresponds  to  P. 

32.  Two  weights,  P  and  Q,  rest  on  the  convex  side  of  a 
smooth  vertical  circle,  and  are  connected  by  a  string  which 
passes  over  a  smooth  peg  vertically  over  the  centre  of  the 
circle  ;  find  the  position  of  equilibrium. 

Ans.  Let  h  =  the  distance  between  the  peg,  B,  and  the 
centre  of  the  circle  ;  6  and  (p  =  the  angles  made  with  the 
vertical  by  the  radii  to  P  and  Q,  respectively  ;  «  and  (3  = 
the  angles  made  with  the  tangents  to  the  circle  at  P  and 
Q  by  the  portions  PB  and  QB  of  the  string;  I  —  length 
of  the  string  ;  then 

p  sin  0  sin  0 

cos  «  ~ "      cos  (3 ' 

7  /sin  6       sin  0\        , 
Vcos  a       cos  (3/ 

h  cos  {6  -f  «)  =  a  cos  cc9 

h  cos  ((f)  -f  |3)  =  a  cos  ft 


CHAPTER     II  I. 

COMPOSITION    AND    RESOLUTION    OF   FORCES   ACTING 
ON    A    RIGID    BODY. 

43.  A  Rigid  Body. — In  the  lust  chapter  wc  considered 
the  action  of  forces  which  have  a  common  point  of  applica- 
tion. We  shall  now  consider  the  action  of  forces  which  are 
applied  at  different  points  of  a  rigid  body. 

A  rigid  body  is  one  in  which  the  particles  retain  invari- 
able positions  with  respect  to  one  another,  so  that  no 
external  force  can  alter  them.  Now,  as  a  matter  of  fact, 
there  is  no  such  thing  in  nature  as  a  body  that  is  perfectly 
rigid  ;  every  body  yields  more  or  less  to  the  forces  which 
act  on  it.  If,  then,  in  any  case,  the  body  is  altered  or  com- 
pressed appreciably,  we  shall  suppose  that  it  has  assumed 
its  figure  of  equilibrium,  and  then  consider  the  points  of 
application  of  the  forces  as  a  system  of  invariable  form. 
The  term  body  in  this  work  means  rigid  body. 

44.  Transmissibility  of  Force. — When  a  force  acts 
at  a  definite  point  of  a  body  and  along  a  definite  line,  the 
effect  of  the  force  will  be  unchanged  at  whatever  point  of 
its  direction  we  suppose  it  applied,  provided  this  point  be 
either  one  of  the  points  of  the  body,  or  be  invariably  con- 
nected with  the  body.  This  principle  is  called  the  trans- 
"nissibility  of  a  force  to  any  point  in  its  line  of  action. 

Now  two  equal  forces  acting  on  a  particle  in  the  same 
line  and  in  opposite  directions  neutralize  each  other  (Art. 
16)  ;  so  by  this  principle  two  equal  forces  acting  in  the 
same  line  and  in  opposite  directions  at  any  points  of  a 
rigid  body  in  that  line  neutralize  each  other.  Hence  it  is 
clear  that  when  many  forces  are  acting  on  a  rigid  body, 
any  two,  which  are  equal  and  have  the  same  line  of  action 


58  RESULTANT  OF  PARALLEL   FORCES. 

and  act  in  opposite  directions,  may  be  omitted,  and  also 
that  two  equal  forces  along  the  same  line  of  action  and  in 
opposite  directions,  may  be  introduced  without  changing 
the  circumstances  of  the  system. 

45.  Resultant  of  Two  Parallel 
Forces.*— (1)  Let  P  and  ft  (Fig. 
14),  be  the  two  parallel  forces  acting 
at  the  points  A  and  B,  in  the  same 
direction,  on  a  rigid  body.  It  is  re- 
quired to  find  the  resultant  of  P 
and  Q. 

At  A  and  B  introduce  two  equal 
and  opposite  forces,  F.  The  introduction  of  these  forces 
will  not  disturb  the  action  of  P  and  Q  (Art.  44).  P  and  F 
at  A  are  equivalent  to  a  single  force,  R,  and  Q  and  F  at  B 
are  equivalent  to  a  single  force,  S.  Then  let  R  and  8  be 
supposed  to  act  at  0,  the  point  of  intersection  of  their  lines 
of  action.  At  this  point  let  them  be  resolved  into  their 
components,  P,  F,  and  Q,  F,  respectively.  The  two  forces, 
F,  at  0,  neutralize  each  other,  while  the  components,  P 
and  Q,  act  in  the  line  OG,  parallel  to  their  lines  of  action 
at  A  and  B.  Hence  the  magnitude  of  the  resultant  is 
P  +  Q,  (Art.  28).  To  find  the  point,  G,  in  which  its  line 
of  action  cuts  AB,  let  the  extremities  of  P  and  R  (acting  at 
A)  be  joined,  and  complete  the  parallelogram.  Then  the 
triangle  PAR  is  evidently  similar  to  GO  A  ;  therefore, 

P       GO  ..    .    Q       GO 

_=_;  similarly  p=gs; 


therefore,  by  division, 

P_  GB 
Q  ~  GA 


(1) 


*  Minchin's  Statics,  p.  85. 


RESULTANT  OF  PARALLEL   FORCES.  59 

(2)  When  tlte  forces  act  in  opposite  directions. — At  A  and 
B,  (Fig  15),  apply  two  equal  and  opposite  forces  F,  as 
before,  and  let  R,  the  resultant  of  P  \  fi 
and  F,  and  S,  the  resultant  of  Q  and 
F,  be  transferred  to  0,  their  point  of 
intersection.  If  at  0  the  forces,  R 
and  S,  are  decomposed  into  their 
original  components,  the  two  forces, 
F,  destroy  each  other,  the  force,  P, 
will  act  in  the  direction  GO  parallel  to  the  direction  of 
P  and  Q,  and  the  force  Q  will  act  in  the  direction  OG. 
Hence  the  resultant  is  a  force  =  P  —  Q,  acting  in  the  line 
GO.  To  find  the  point  G,  we  have,  from  the  similar 
triangles,  PAR  and  OGA, 

F~  GA'  amF-<m> 

Hence  the  resultant  of  two  parallel  forces,  acting  in  the 
same  or  opposite  directions,  at  the  extremities  of  a  rigid 
right  line,  is  parallel  to  the  components,  equal  to  their 
algebraic  sum,  and  divides  tjie  line  or  the  line  produced, 
into  two  segments  which  are  inversely  as  the  forces. 

In  both  cases  we  have  the  equation 

P  x  GA  =  Q  x  GB.  (3) 

Hence  the  following  theorem : 

If  from  a  point  on  the  resultant  of  two  parallel  forces  a 
right  line  be  drawn  meeting  the  forces,  luhether  perpendicu- 
larly or  not,  the  products  obtained  by  multiplying  each  force 
by  its  distance  from  the  resultant,  measured  along  the  arbi- 
trary line,  are  equal. 

Sch. — The  point  G  possesses  this  remarkable  property ; 


60  MOMENT    OF   A    FORCE. 

that,  however  P  and  Q  are  turned  about  their  points  of 
application,  A  and  B,  their  directions  remaining  parallel, 
G,  determined  as  above,  remains  fixed.  This  point  is  in 
consequence  called  the  centre  of  the  parallel  forces,  P 
and  Q. 

46.  Moment  of  a  Force. — The  moment  of  a  force  ivith 
respect  to  a  point  is  the  product  of  the  force  and  the  perpen- 
dicular let  fall  on  its  line  of  action  from  the  point.  The 
moment  of  a  force  measures  its  tendency  to  produce  rota- 
tion about  a  fixed  point  or  fixed  axis. 
Thus  let  a  force,  P,  (Fig.  16),  act  on 
a  rigid  body  in  the  plane  of  the  paper, 
and  let  an  axis  perpendicular  to  this 
plane  pass  through  the  body  at  any 
point,  0.  It  is  clear  that  the  effect  of 
the  force  will  be  to  turn  the  body  round  this  axis  (the  axis 
being  supposed  to  be  fixed),  and  the  turning  effect  will 
depend  on  the  magnitude  of  the  force,  P,  and  the  perpen- 
dicular distance,  p,  of  P  from  0.  If  P  passes  through  0, 
it  is  evident  that  no  rotation  of  the  body  round  0  can  take 
place,  whatever  be  the  magnitude  of  P ;  while  if  P 
vanishes,  no  rotation  will  take  place  however  great  p  may 
be.  Hence,  the  measure  of  the  power  of  the  force  to 
produce  rotation  may  be  represented  by  the  product 

p-p, 

and  this  product  has  received  the  special  name  of  Moment. 

The  unit  of  force  being  a  pound  and  the  unit  of  length  a 
foot,  the  unit  of  moment  will  evidently  be  a,  foot-pound. 

The  point  0  is  called  the  origin  of  moments,  and  may  or 
may  not  be  chosen  to  coincide  with  the  origin  of  co- 
ordinates. The  solution  of  problems  is  often  greatly  sim- 
plified by  a  proper  selection  of  the  origin  of  moments.  The 
perpendicular  from  the  origin  of  moments  to  the  action  line 
of  the  force  is  called  the  arm  of  the  force. 


SIGNS    OF   MOMENTS.  61 

47.  Signs  of  Moments. — A  force  may  tend  to  turn  a 
body  about  a  point  or  about  an  axis,  in  either  of  two  direc- 
tions ;  if  one  be  regarded  as  positive  the  other  must  be 
negative ;  and  hence  we  distinguish  between  positive  and 
negative  moments.  For  the  sake  of  uniformity  the  moment 
of  a  force  is  said  to  be  negative  when  it  tends  to  turn  a  body 
from  left  to  right,  i.  <?.,  in  the  direction  in  which  the  hands 
of  a  clock  move  ;  and  positive  when  it  tends  to  turn  the 
body  from  right  to  left,  or  opposite  the  direction  in  which 
the  hands  of  a  clock  move. 

48.  Geometric  Representation  of  the  Moment  of 
a  Force  with  respect  to  a  Point. — Let  the  line  AB 
(Fig.  16),  represent  the  force,  P,  in  magnitude  and  direc- 
tion, and^  the  perpendicular  OC  ;  then  the  moment  of  P 
with  respect  to  0  is  AB  xp  (Art.  46).  But  this  is  double 
the  area  of  the  triangle  AOB.  Hence,  the  moment  of  a  force 
with  respect  to  a  point  is  geometrically  represented  by  double 
the  area  of  the  triangle  whose  base  is  the  line  representing 
the  force  in  magnitude  and  direction,  and  ivhose  vertex  is 
the  given  point. 

49.  Case  of  Two  Equal  and  Opposite  Parallel 
Forces. — If  the  forces,  P  and  Q,  in  Art.  45,  (Fig.  15)  are 
equal,  the  equation 

P  x  GA  =  Q  x  GB 

gives  GA  =  GB,  which  is  true  only  wThen  G  is  at  infinity 
on  AB ;  also  the  resultant,  P — Q,  is  equal  to  zero.  Such  a 
system  is  called  a  Couple. 

A  Couple  consists  of  two  equal  and  opposite  parallel  forces 
acting  on  a  rigid  body  at  a  finite  distance  from  each  other. 

We  shall  investigate  the  laws  of  the  composition  and 
resolution  of  couples,  since  to  these  the  composition  and 


62  MOMENT    OF   A     COUPLE. 

resolution  of  forces  of  every  kind  acting  on  a  rigid  body 
may  be  reduced. 


6     d 


50.  Moment  of  a  Couple. — Let  0  f 
(Fig.  17)  be  any  point  in  the  plane  of  the 
couple ;  let  fall  the  perpendiculars  Oa 
and  Ob  on  the  action  lines  of  the  forces 
P.  Then  if  0  is  inside  the  lines  of  action 
of  the  forces,  both  forces  tend  to  produce  F'9'17 

rotation  round  0  in  the  same  direction,  and  therefore  the 
sum  of  their  moments  is  equal  to 

P  (Oa  +  Ob),  or  P  x  ab 

If  the  point  chosen  is  0',  the  sum  of  the  moments  is 
evidently 

P  (0'«  _  O'b),  or  P  x  ab, 

which  is  the  same  as  before.  Hence  the  moment  of  the 
couple  with  respect  to  all  points  in  its  plane  is  constant. 

The  Arm  of  a  couple  is  the  perpendicular  distance 
between  the  two  forces  of  the  couple. 

The  Moment  of  a  couple  is  the  product  of  the  arm  and 
one  of  the  forces. 

The  Axis  of  a  couple  is  a  right  line  drawn  from  any 
chosen  point  perpendicular  to  the  plane  of  the  couple,  and 
of  such  length  as  to  represent  the  magnitude  of  the  mo- 
ment, and  in  such  direction  as  to  indicate  the  direction  in 
which  the  couple  tends  to  turn. 

As  the  motion,  in  Statics  is  only  virtual,  and  not  actual, 
the  direction  of  the  axis  is  fixed,  but  not  the  position  of  it; 
it  may  be  any  line  perpendicular  to  the  plane  of  the  couple, 
and  may  be  drawn  as  follows;  imagine  a  watch  placed  in 
the  plane  in  which  several  couples  act.  Then  let  the  axes 
of  those  couples  which  tend  to   produce  rotation   in   the 


COUPLES.  63 

direction  of  the  motion  of  the  hands  be  drawn  downward 
through  the  back  of  the  watch,  and  the  axes  of  those  which 
tend  to  produce  the  contrary  rotation  be  drawn  upward 
through  the  face  of  the  watch.  Thus  each  couple  is  com- 
pletely represented  by  its  axis,  which  is  drawn  upward  or 
downward  according  as  the  moment  of  the  couple  is  posi- 
tive or  negative ;  and  couples  are  to  be  resolved  and 
compounded  by  the  same  geometric  constructions  performed 
with  reference  to  their  axes  as  forces  or  velocities,  with 
reference  to  the  lines  which  directly  represent  them. 

We  shall  now  give  three  propositions  showing  that  the 
effect  of  a  couple  is  not  altered  when  certain  changes  are 
made  with  respect  to  the  couple. 

51.  The  Effect  of  a  Couple  on  a  Rigid  Body  is  not 
altered  if  the  arm  be  turned  through  any  angle 
about  one  extremity  in  the  plane  of  the  Couple. 

Let  the  plane  of  the  paper  be  •  the 
plane  of  the  couple,  AB  the  arm  of 
the  original  couple,  AB'  its  new  posi- 
tion, and  P,  P,  the  forces.  At  A 
and  B'  respectively  introduce  two 
forces  each  equal  to  P,  with  their 
action  lines  perpendicular  to  the  arm 
AB',  and  opposite  in  direction  to 
each  other.  The  effect  of  the  given 
couple  is,  of  course,  unaltered  by  the  introduction  of  these 
forces.  Let  BAB'  =  20  ;  then  the  resultant  of  P  acting  at 
B,  and  of  P  acting  at  B',  whose  lines  of  action  meet  at  Q> 
is  2P  sin  0,  acting  along  the  bisector  AQ;  and  the  result- 
ant of  P  acting  at  A  perpendicular  to  AB  and  of  P  per- 
pendicular to  AB',  is  2P  sin  0,  acting  along  the  bisector 
AQ  m  &  direction  opposite  to  the  former  resultant.  Hence 
these  two  resultants  neutralize  each  other ;  and  there 
remains  the  couple  whose  arm  is  AB',  and  whose  forces  are 
P,  P.     Hence  the  effect  of  the  couple  is  not  altered. 


64 


COUPLES. 


r 


Fig. 19 


52.  The  Effect  of  a  Couple  on  a  Rigid  Body  is 
not  filtered  if  we  transfer  the  Couple  to  any  at  Iter 
Parallel  Plane,  the  Ann  remaining  parallel  to 
itself. 

Let  AB  be  the  arm,  and  P,  P,  the  f 
forces  of  the  given  couple;  let  A'B' 
be  the  new  position  of  the  arm  par- 
allel to  AB.  At  A'  and  B'  apply  two 
equal  and  opposite  forces  each  equal 
to  P,  acting  perpendicular  to  A'B', 
and  in  a  plane  parallel  to  the  plane  of 
the  original  couple.  This  will  not  alter  the  effect  of  the 
given  couple.  Join  AB',  A'B,  bisecting  each  other  at  0  ; 
then  P  at  A  and  P  at  B',  acting  in  parallel  lines,  and  in 
the  same  direction,  are  equivalent  to  2P  acting  at  0  ;  also 
P  at  B  and  P  at  A',  acting  in  parallel  lines  and  in  the 
same  direction,  are  equivalent  to  2P  acting  at  0.  At  0 
therefore  these  two  resultants,  being  equal  and  opposite, 
neutralize  each  other ;  and  there  remains  the  couple  whose 
arm  is  A'B',  and  whose  forces  are  each  P,  acting  in  the 
same  directions  as  those  of  the  original  couple.  Hence  the 
effect  of  the  couple  is  not  altered. 

53.  The  Effect  of  a  Couple  on  a  Rigid  Body  is 
not  altered  if  we  replace  it  by  another  Couple  of 
which  the  Moment  is  the  same;  the  Plane  rem  (till- 
ing the  same  and,  the  Arms  being  in  the  same 
straight  line  and  having  a 
common  extremity. 

Let  AB  be  the  arm,  and  P,  P,  the 
forces  of  the  given  couple,  and  sup- 
pose P  =  Q  -f-  R.    Produce  AB  to  C  F!g,26 
so  that  1p=q+r 
AB  :  AC    ::     Q  :  P  (=  Q  +  R), 


P=Q+R 


and  therefore        AB   :   BC 


Q   :    R\ 


(1) 
(2) 


FORCE    AND    A     COUPLE.  66 

at  C  introduce  opposite  forces  each  equal  to  Q  and  parallel 
to  P  ;  this  will  not  alter  the  effect  of  the  couple. 

Now  R  at  A  and  Q  at  C  will  halance  Q  +  R  at  B  froin 
(2)  and  (Art.  45) ;  hence  there  remain  the  forces,  Q,  Q, 
acting  on  the  arm,  AC,  which  form  a  couple  whose  moment 
A  equal  to  that  of  Pt  P,  with  arm,  AB,  since  by  (1)  we 
have 

P  x  AB  =  Q  x  AC. 

Hence  the  effect  of  the  couple  is  not  altered. 

Rem. — From  the  last  three  articles  it  appears  that  we 
may  change  a  couple  into  another  couple  of  equal  moment, 
and  transfer  it  to  any  position,  either  in  its  own  plane  or 
in  a  plane  parallel  to  its  own,  without  altering  the  effect  of 
the  couple.  The  couple  must  remain  unchanged  so  far  as 
concerns  the  direction  of  rotation  which  its  forces  would 
tend  to  give  the  arm,  i.  e.,  the  axis  of  the  couple  may  be 
removed  parallel  to  itself,  to  any  position  within  the  body 
acted  on  by  the  couple,  while  the  direction  of  the  axis  from 
the  plane  of  the  couple  is  unaltered  (Art.  50). 

54.  A  Force  and  a  Couple  acting  in  the  same 
Plane  on  a  Rigid  Body  are  equivalent  to  a  Single 
Force. 

Let  the  force  be  F  and  the  couple  (P,  a),  that  is,  P  is 
the  magnitude  of  each  force  in  the  couple  whose  arm  is  a. 

Then  (Art.  53)  the  couple  (P,  a)  =  the  couple  If,  -^y 

Let  this  latter  couple  be  moved  till  one  of  its  forces  acts  in 
the  same  line  as  the  given  force,  F,  but  in  the  opposite 
direction.  The  given  force,  F,  will  then  be  destroyed,  and 
there  will  remain  a  force,  F,  acting  in  the  same  direction 
as  the  given  one  and  at  a  perpendicular  distance  from  it 
_aP 


66  RESULTANT    OF    COUPLES. 

Cor. — A  force  and  a  couple  acting  on  a  rigid  body  cannot 
produce  equilibrium.  A  couple  can  be  in  equilibrium  only 
with  an  equivalent  couple.  Equivalent  couples  are  those 
tvhose  moments  are  equal,* 

The  resultant  of  several  couples  is  one  which  will  produce 
the  same  effect  singly  as  the  component  couples. 

55.  To  find  the  Resultant  of  any  number  of 
Couples  acting  on  a  Body,  the  Planes  of  the 
Couples  being  parallel  to  each  other. 

Let  P,  Q,  R,  etc.,  be  the  forces,  and  a,  b,  c,  etc.,  their 
arms  respectively.  Suppose  all  the  couples  transferred  to 
the  same  plane  (Art.  52) ;  next,  let  them  all  be  transferred  so 
as  to  have  their  arms  in  the  same  straight  line,  and  one 
extremity  common  (Art.  51)  ;  lastly,  let  them  be  replaced 
by  other  couples  having  the  same  arm  (Art.  53).  Let  a  be 
the  common  arm,  and  Pu  Q19  Rt,  etc.,  the  new  forces, 
so  that 

Pta  =  Pa,     Qxa  =  Qb,    Rta  —  Re,  etc., 
then       Px  =  P-,    Qt  =  Q-9   Rt  =  R-,  etc., 

i.  e.,  the  new  forces  are  P-,  Q-,  R-,  etc.,  acting  on  the 

common  arm  a.     Hence  their  resultant  will  be  a  couple  of 
which  each  force  equals 

and  the  arm  =  cc,  or  the  moment  equals 
Pa  +  Qb  +  Re  +  etc. 
If  one  of  the  couples,  as  Q,  act  in  a  direction  opposite  to 

*  The  moments  of  equivalent  couples  may  have  like  or  unlike  signs 


RESULTANT    OF    TWO     COUPLES.  67 

the  other  couples  its  sign  will  be  negative,  and  the  force  at 
each  extremity  of  the  arm  of  the  resultant  couple  will  be 

P--Qh  +  JSl  +  etc 

Hence  the  moment  of  the  resultant  couple  is  equal  to  the 
algebraic  sum  of  the  moments  of  the  component  couples. 

56.  To  Find  the  Resultant  of  two  Couples  not 
acting  in  the  same  Plane* 

Let  the  planes  of  the  couples  be 
inclined  to  each  other  at  an 
angle  y  ;  let  the  couples  be  trans- 
ferred in  their  planes  so  as  to 
have  the  same  arm  lying  along 
the  line  of  intersection  of  the  two 
planes  ;  and  let  the  forces  of  the 

couples  thus  transferred  be  P  and  Q.  Let  AB  be  the  com- 
mon arm.  Let  R  be  the  resultant  of  the  forces  P  and  Q  at 
A  acting  in  the  direction  AR  ;  and  of  P  and  Q  at  B  acting 
in  the  direction  BR.  Then  since  P  and  Q  at  A  are  parallel 
to  P  and  Q  at  B  respectively,  therefore  R  at  A  is  parallel 
to  R  at  B.  Hence  the  two  couples  are  equivalent  to  the 
single  couple  R,  R,  acting  on  the  arm  AB  ;  and  since 
PAQ  =  y,  we  have 

R2  =  P2  +  Q2  +  2PQ  cos  y  (Art.  30).  (1) 

Draw  Aa,  Bb  perpendicular  to  the  planes  of  the  couples 
P,  P,  and  ft  Qy  respectively,  and  proportional  in  length  to 
their  moments. 

Draw  Kg  perpendicular  to  the  plane  of  R,  i?,  and  in  the 
same  proportion  to  A#,  Bb,  that  the  moment  of  the  couple, 
R,  R,  is  to  those  of  P,  P,  and  Q,  Q,  respectively.  Then 
Aa,  Kb,  Ac,  may  be  taken  as  the  axes  of  P,  P ;  Q,  Q ;  and 

*  Todhunter's  Statics,  p.  42.    Also  Pratt's  Mechanics,  p.  25- 


d8  resultant  of  two   couples. 

R,  R,  respectively  (Art.  50).  Now  the  three  straight  lines, 
Aa,  Ac,  Ab,  make  the  same  angles  with  each  other  that 
AP,  AR,  AQ  make  with  each  other;  also  they  are  in  the 
same  proportion  in  which 

AB  •  P,    AB  •  R,    AB  .  Q  are, 

or  in  which  P,  R,  Q  are. 

But  R  is  the  resultant  of  P  and  Q  ;  therefore  Ac  is  the 
diagonal  of  the  parallelogram  on  Aa,  Ab  (Art.  30). 

Hence  if  two  straight  lines,  having  a  common  extremity, 
represent  the  axes  of  two  couples,  that  diagonal  of  the 
parallelogram  described  on  these  straight  lines  as  adjacent 
sides  which  passes  through  their  common  extremity  repre- 
senls  the  axis  of  the  resultant  couple. 

Cor. — Since  R  •  AB  is  the  axis  or  moment  of  the  result- 
ant couple,  we  have  from  (1) 

R*.  AB2  =  P2- AB24-§2- AB2+2P-  AB-  ^-AB-cosy.  (2) 

If  L  and  M  represent  the  axes  or  moments  of  the  com- 
ponent couples  and  G,  that  of  the  resultant  couple,  (2) 
becomes 

GP  =  I?  +  if2  +  %L  •  ^cos  y.  (3) 

Sch.  1. — If  L,  M,  N,  are  the  axes  of  three  component 
couples  which  act  in  planes  at  right  angles  to  one  another, 
and  G  the  axis  of  the  resultant  couple,  it  may  easily  be 
shown  that 

G*  =  I?  +  M 2  +  N\  (4) 

If  A,  n,  v  be  the  angles  which  the  axis  of  the  resultant 
makes  with  those  of  the  components,  we  have 

L  M  N 

COS  A  =  — ,       COS  fJL  =   77,       COS   V  =  -jr' 
Cr  Cr  Cr 


VARIGXOX'S    THEOREM    OF   MOMENTS,  6(J 

Sch.  2. — Hence,  conversely  any  couple  may  be  replaced 
by  three  couples  acting  in  planes  at  right  angles  to  one 
another ;  their  moments  being  (7  cos  A,  G  cos  \i,  G  cos  v  • 
where  G  is  the  moment  of  the  given  couple,  and  A,  //,  v  the 
angles  its  axis  makes  with  the  axes  of  the  three  couples. 

Thus  the  composition  and  resolution  of  couples  follow 
laws  similar  to  those  which  apply  to  forces,  the  axis  of  the 
couple  corresponding  to  the  direction  of  the  force,  and  the 
moment  of  the  couple  to  the  magnitude  of  the  force. 

57.  Varignon's  Theorem  of  Moments. — The  mo- 
ment of  the  resultant  of  two  component  forces 
with  respect  to  any  point  in  their  plane  is  equal 
to  the  algebraic  sum  of  the  moments  of  the  two 
components  with  respect  to  the  same  point. 

Let  A  P  and  A  Q  represent  two  com- 
ponent forces ;  complete  the  parallelo- 
gram and  draw  the  diagonal,  A R, 
representing  the  resultant  force.  Let 
0  be  the  origin  of  moments  (Art.  46). 
Join  OA,  OP,  OQ,  OR,  and  draw  PC  '9' 

and  QB  parallel  to  OA,  and  let  p  =  the  perpendicular  let 
fall  from  0  to  AR. 

Now  the  moment  of  AP  about  0  is  the  product  of  AP 
and  the  perpendicular  let  fall  on  it  from  0  (Art.  46),  which 
is  double  the  area  of  the  triangle,  A  OP  (Art.  48).  But 
the  area  of  the  triangle,  A  OP,  =  the  area  of  the  triangle, 
AOC,  since  these  triangles  have  the  same  base,  AO,  and 
are  between  the  same  parallels,  AO  and  CP.  Hence  the 
moment  of  AP  about  0  =  the  moment  of  AC  about 
0  =  AC- p.  Also  the  moment  of  AQ  about  0  is  double 
the  area  of  the  triangle,  A  OQ,  =  double  the  area  of  the 
triangle,  A  OB,  since  the  two  triangles  have  the  same  base, 
AO,  and  are  between  the  same  parallels,  AO  and  QB, 
Hence  the  moment  of  AQ  about  0  =  the  moment  of  AB 


70  VARIGNON'S  theorem  of  moments. 

about  0  =  AB  •  p.  Therefore  the  sum  of  the  moments  oi 
AP  and  AQ  about  0  =  the  sum  of  the  moments  of  AC 
and  AB  about  0  =  (AC  +  -45) i>,  =  (AB  +  £i2)jt?, 
(since  ^iC  =  BR  from  the  equal  triangles  APC  and  QBE) 
=  AB  •  p  =  the  moment  of  the  resultant. 

If  the  origin  of  moments  fall  between  AP  and  AQ,  the 
forces  will  tend  to  produce  rotation  in  opposite  directions, 
and  hence  their  moments  will  have  contrary  signs  (Art. 
47).  In  this  case  the  moment  of  the  resultant  =  the  dif- 
ference of  the  moments  of  the  components,  as  the  student 
will  find  no  difficulty  in  showing.  Hence  in  either  case 
the  moment  of  the  resultant  is  equal  to  the  algebraic  sum 
of  the  moments  of  the  components. 

Cor.  1. — If  there  are  any  number  of  component  forces, 
we  may  compound  them  in  order,  taking  any  two  of  them 
first,  then  finding  the  resultant  of  these  two  and  a  third, 
aad  so  on ;  and  it  follows  that  the  sum  of  their  moments 
(with  their  proper  signs),  is  equal  to  the  moment  of  the 
resultant. 

Cor.  2. — If  the  origin  of  moments  be  on  the  line  of 
action  of  the  resultant,  p  =  0,  and  therefore  the  moment 
of  the  resultant  =  0  ;  hence  the  sum  of  the  moments  of 
the  components  is  equal  to  zero.  In  this  case  the  moments 
of  the  forces  in  one  direction  balance  those  in  the  opposite 
direction ;  i.  e.,  the  forces  that  tend  to  produce  rotation  in 
one  direction  are  counteracted  by  the  forces  that  tend  to 
produce  rotation  in  the  opposite  direction,  and  there  is  no 
tendency  to  rotation. 

Cor.  3. — If  all  the  forces  are  in  equilibrium  the  resultant 
R  =  0,  and  therefore  the  moment  of  R  =  0 ;  hence  the 
sum  of  the  moments  of  the  components  is  equal  to  zero, 
and  there  is  no  tendency  to  motion  either  of  translation  or 
rotation. 


VARIGXOX'S   THEOREM  FOR   PARALLEL   FORCES.      71 

Cor.  4. — Therefore  when  the  moment  of  the  resultant 
=  0,  we  conclude  either  that  the  resultant  =  0  (Cor.  3), 
or  that  it  passes  through  the  point  taken  as  the  origin  of 
moments  (Cor.  2). 

58.  Varignon's  Theorem  of  Moments  for  Parallel 
Forces. — The  sum  of  the  moments  of  two  parallel 
forces  about  any  point  is  equal  to  the  moment  of 
their  resultant  about  the  point. 


Let  P  and  Q  be  two  parallel  forces 
acting  at  A  and  B,  and  R  their  result- 
ant acting  at  G,  and  let  0  be  the  point 
about  which  moments  are  to  be  taken. 
Then  (Art.  45)  we  have 


Fig.23 


P  x  AG  =  Q  x  BG, 
.-.    P(0G  -  OA)  =  Q  (OB  -  OG), 
.'.       (P  +  Q)  OG  =  P  x  OA  +  Q  x  OB, 

i?xOG  =  PxOA+exOB; 

that  is,  the  sum  of  the  moments  =s  the  moment  of  the 
resultant. 

Cor. — It  follows  that  the  algebraic  sum  of  the  moments 
of  any  number  of  parallel  forces  in  one  plane,  with  respect 
to  a  point  in  their  plane,  is  equal  to  the  moment  of  their 
resultant  with  respect  to  the  point. 

59.  Centre  of  Parallel  Forces. — To  find  the  mag- 
nitude, direction,  and  point  of  application  of  the 
resultant  of  any  number  of  parallel  forces  acting 
vn  a  rigid  body  in  one  plane. 


72 


CENTRE   OF  PARALLEL   FORCES. 


2> 


Let  Px,  P2,  P3,  etc.,  denote  the 
forces,  M1%  Jf2,  M3,  etc.,  their  points 
of  application.  Take  any  point  in 
the  plane  of  the  forces  as  origin  and 
draw  the  rectangular  axes  OX,  OY. 
Let  (xx,  yx),  (x2,  y2),  etc.,  be  the 
points   of  application,   Mu   M2,    etc.  F'g.24 

Join  MXM2\  and  take  the  point  if  on  MXM2,  so  that 


MXM  P2 

MXM2  ~  Pt  +P,; 


(1) 


tlien  the  resultant  of  Px  and  P2  is  Pj  +  P2,  and  it  acts 
through  M  parallel  to  Px  (Art.  45). 

Draw  Mxa,  Mb,  M2c  parallel,  and  Mxe  perpendicular  to 
the  axis  of  y.    Then  we  have 

MXM     _  Md  __  Mb  —  yx 


MXM< 


M9 


Pc 


Mb  -  yt  =  p-^rp-  {y*  ~ 

m  _  p^x  +  P%y% . 
-    -     A  +  p2    ' 


yj; 


(2) 


which  gWes  the  ordinate  of  the  point  of  application  of  the 
resultant  pf  Px  and  P2. 

Now  since  the  resultant  of  Px  and  P2,  which  is 
Px  -f  P2,  acts  at  M,  the  resultant  of  Px  -f  P2  at  if,  and 
P3  at  i¥3,  is  Px  +  P3  4-  P3  at  g,  and  substituting  in  (2) 
Px  +  P2,  P3,  Jf£,  and  yz  for  P1?  P2,  yx,  and  ?/2  respec- 
tively,  we  have 


gk  =: 


(Pt  +  P2)  Mb  +  P,y3  _  P^t+P^+Ps.^ 


A  +  A  +  *, 


P.+P2  +  P3 


;(3) 


CENTRE   OF  PARALLEL   FORCES.  73 

and  this  process  may  be  extended  to  any  number  of  parallel 
forces.  Let  R  denote  the  resultant  force  and  y  the  ordi- 
nate of  the  point  of  application  ;  then  we  have 

R  =  Px  +  P3  +  R3 .  +  etc.  =  SP. 


7/  -  ^  +  PM*  +  PsVs  +  etc.        £P^ 
J  "  Pt  -f-  Ps  +  P3  +  etc.  "   2P  ' 

Similarly,  if  a?  be  the  abscissa  of  the  point  of  application  o^ 
the  resultant,  we  have 

ZPx 

x=Yp' 

The  values  of  x,  y  are  independent  of  the  angles  which 
the  directions  of  the  forces  make  with  the  axes.  Hence 
if  these  directions  be  turned  about  the  points  of  application 
of  the  forces,  their  parallelism  being  preserved,  the  point  of 
application  of  the  resultant  will  no't  move.  For  this  reason 
the  point  (x,  y)  is  called  the  centre  of  parallel  forces.  We 
shall  hereafter  have  many  applications  in  which  its  position 
is  of  great  importance. 

Sch.  1. — The  moment  of  a  force  with  respect  to  a  plane 
is  the  product  of  the  force  into  the  perpendicular  distance 
of  its  point  of  application  from  the  plane.  Thus,  Pxyx  is 
the  moment  of  the  force  Px,  in  reference  to  the  plane 
through  OX  perpendicular  to  OY.  This  must  be  carefully 
distinguished  from  the  moment  of  a  force  with  respect  to 
a  point.  Hence  the  equations  for  determining  the  position 
of  the  centre  of  parallel  forces  show  that  the  sum  of  the 
moments  of  the  parallel  forces  with  respect  to  any  plane,  is 
equal  to  the  moment  of  their  resultant. 

Sch.  2. — The  moment  of  a  force  with  respect  to  any  line 
is  the  product  of  the  component  of  the  force  perpendicular 
4 


Fig.25 


74  CONDITIONS   OF  EQUILIBRIUM. 

to  the  line  into  the  shortest  distance  between  the  line  and 
the  line  of  action  of  the  force. 

60.   Conditions  of  Equilibrium  of  a  Rigid  Body 
acted   on   by   Parallel   Forces   in  one  Plane. — Let 

Plf  P2,  P3,  etc.,  denote  the  forces.  Take 
any  point  in  the  plane  of  the  forces  as 
origin,  and  draw  rectangular  axes,  OX, 
OY,  the  latter  parallel  to  the  forces.  Let 
A  be  the  point  where  OX  meets  the  direc- 
tion of  Plf  and  let  OA  =  xx. 

Apply  at  0  two  opposing  forces,  each  T-fs 
equal  and  parallel  to  P1 ;  this  will  not  disturb  the  equili- 
brium. Then  P±  at  A  is  replaced  by  Pt  at  0  along  OY, 
and  a  couple  whose  moment  is  Px  •  OA,  i.  e.,  P^.  The 
remaining  forces,  P2,  P3,  etc.,  may  be  treated  in  like  man- 
ner. We  thus  obtain  a  set  of  forces,  Px,  P2,  P3,  etc., 
acting  at  0  along  OY,  and  a  set  of  couples,  Pxxx,  P2x2, 
P3x3,  etc.,  in  the  plane  of  the  forces  tending  to  turn  the 
body  from  the  axis  of  #  to  the  axis  of  y.  These  forces  are 
equivalent  to  a  single  resultant  force  Px  +  P2  +  P3  -f  etc., 
and  the  couples  are  equivalent  to  a  single  resultant  couple, 
Ptxt  +  P2x2  +  P.dxz  +  etc.  (Art.  55). 

Hence  denoting  the  resultant  force  by  R,  and  the  moment 
of  the  resultant  couple  by  G,  we  have 

R  =  Pt  +  P2  +  P3  +  etc.  =  2P; 

G  =  P1x1  +  P2x2  +  Pzxz  +  etc.  =  ZPx; 

that  is,  a  system  of  parallel  forces  can  be  reduced  to  a 
single  force  and  a  couple,  which  (Art.  54,  Cor.)  cannot 
produce  equilibrium.  Hence,  for  equilibrium,  the  force 
and  the  couple  must  vanish  ;  or 

ZP  =  0,     and     ZPx  =  0. 


y 

Yd  * 

ft 

X 

u 

,  A 

x 

,v. 

0 

M 
Fig.26 

CONDITIONS    OF  EQUILIBRIUM.  75 

Hence  the  conditions  of  equilibrium  of  a  system  of  par- 
allel forces  acting  on  a  rigid  body  in  one  plane  are  : 

The  sum  of  the  forces  must  =  0. 

Tlie  sum  of  the  moments  of  the  forces  about  every  point  in 
their  plane  must  =  0. 

61.  Conditions  of  Equilibrium  of  a  Rigid  Body- 
acted  on  by  Forces  in  any  direction  in  one  Plane. — 

Let  P19  P2,  P3,  etc.,  be  the  forces  acting  at  the  points 

(*ii  !fi)> •(*•»  y2)>  (xs>  Vz)y  etc->  in  the 
plane  xy.  Resolve  the  force  Pt  into  two 
components,  X19  Y19  parallel  to  OX 
and  OY  respectively.  Let  the  direc- 
tion of  Y1  meet  OX  at  M,  and  the 
direction  of  Xx  meet  OY  at  N.  Apply 
at  0  two  opposing  forces  each  equal  and  parallel  to  Xl9 
and  also  two  opposing  forces  each  equal  and  parallel  to  Yt . 
Hence  Y1  at  A i9  or  M,  is  equivalent  to  Yt  at  0,  and  a 
couple  whose  moment  is  Yt  •  OM ;  and  Xt  at  A19  or  JV,  is 
equivalent  to  Xx  at  0,  and  a  couple  whose  moment  is 
Xt  ■  ON. 

Hence  Yt  is  replaced  by  Yj  at  0,  and  the  couple  Yxzt  ; 
and  Xt  is  replaced  by  Xt  at  0,  and  the  couple  Xxyx  (Art. 
47).  Therefore  the  force  Pt  may  be  replaced  by  the  com- 
ponents X19  Yt  acting  at  0,  and  the  couple  whose 
moment  is 

Yxxx  -  X,y19 

and  which  equals  the  moment  of  Pt  about  0  (Art.  57). 

By  a  similar  resolution  of  all  the  forces  we  shall  have 
them  replaced  by  the  forces  (X2,  Y2),  (X3,  Y3)9  etc., 
acting  at  0  along  the  axes,  and  the  couples 

■*  2^2  -^-2^2?        -*   3*^3  -^-3^3?    ^^* 

Adding  together  the  couples  or  moments  of  P2,  P3,  etc., 


76  EQUILIBRIUM    UNDER     THREE    FORCES. 

and  denoting  by  G  the  moment  of  the  resultant  couple,  we 
get  the  total  moment 

e^S(Fx-Xy). 

If  the  sum  of  the  components  of  the  forces  along  OX  is 
denoted  by  HZ,  and  the  sum  of  the  components  along  OT 
by  Ii  Y,  the  resultant  of  the  forces  acting  at  0  is  given  by 
the  equation 

R?=z  (2X)2  +  (S7)«. 

If  a  be  the  angle  which  R  makes  with  the  axis  of  X,  we 
aave 

R  cos  a  =  SJ,     R  sin  a  =  ZYi 

.•.    tan  a  =  —g 

Therefore,  any  system  of  forces  acting  in  any  direction 
in  one  plane  on  a  rigid  body  may  be  reduced  to  a  single 
force,  R,  and  a  single  couple  whose  moment  is  G,  which 
(Art.  54,  Cor.)  cannot  produce  equilibrium.  Hence  for 
equilibrium  we  must  have  R  =  0,  and  '  G  =  0,  which 
requires  that 

SX=  0,    27=  0, 

Z(rx  —  Xy)  -  0. 

Hence  the  conditions  of  equilibrium  for  a  system  of 
forces  acting  in  any  direction  in  one  plane  on  a  rigid  body 
are  : 

The  sum  of  the  components  of  the  forces  parallel  to  each  of 
two  rectangular  axes  must  =  0. 

TJie  sum  of  the  moments  of  the  forces  round  every  point  h 
their  plane  must  =  0 


EXAMPLES.  77 

Cor. — Conversely,  if  the  forces  are  in  equilibrium  the 
sum  of  the  components  of  the  forces  parallel  to  any  direc- 
tion will  ±=  0,  aud  also  the  sum  of  the  moments  of  the 
forces  about  any  point  will  =  0. 

62.  Condition  of  Equilibrium  of  a  Body  under  the 
Action  of  Three  Forces  in  one  Plane. — //  three 
forces  maintain  a  body  in  equilibrium,  their 
directions  must  meet  in  a  point,  or  be  parallel* 

Suppose  the  directions  of  two  of  the  forces,  P  and  Q,  to 
meet  at  a  point,  and  take  moments  round  this  point ;  then 
the  moment  of  each  of  these  two  forces  =  0 ;  therefore  the 
moment  of  the  third  force  R  =  0  (Art.  61,  Cor.),  which 
requires  either  that  R  =  0,  or  that  it  pass  through  the 
point  of  intersection  of  P  and  Q.  If  R  is  not  =  0,  it  must 
pass  through  this  point.  Hence  if  any  two  of  the  forces 
meet,  the  third  must  pass  through  their  point  of  intersec- 
tion, and  keep  it  at  rest,  and  each  force  must  be  equal  and 
opposite  to  the  resultant  of  the  other  two.  If  the  angles 
between  them  in  pairs  be  p,  q}  r,  the  forces  must  satisfy  the 
conditions 

P  :  Q  :  R  =  sin  p  :  sin  q  :  sin  r  (Art.  32). 

If  two  of  the  forces  are  parallel,  the  third  must  be 
parallel  to  them,  and  equal  and  directly  opposed  to  their 
resultant. 

EXAM  PLES. 

2.  Suppose  six  parallel  forces  proportional  to  the  numbers 
1,  2,  3,  4,  5,  6  to  act  at  points  (  —  2,  —1),  (  —  1,  0),  (0,  1), 
(1,  2),  (2,  3),  (3,  4)  ;  find  the  resultant,  R9  and  the  centre 
of  parallel  forces. 

By  Art.  59  we  have 

i2  =  2P  =  l-f2-f...6:=21; 


78  EXAMPLES. 

ZPx  =  _2  —  2  +  4  +  10  +  18  =  28; 
ZPy  =  —  1  +  3  +  8  +  15  +  24  =  49. 

•'•    x  -    ZP    —  21;    y  ""    £P   ~  21* 

2.  At  the  three' vertices  of  a  triangle  parallel  forces  are 
applied  which  are  proportional  respectively  to  the  opposite 
sides  of  the  triangle ;  find  the  centre  of  these  forces. 

Let^j,  yx),  (x2,  y2),  (x3,  yz)  be  the  vertices,  and  let  a, 
b9  c  be  the  sides  opposite  to  them ;  then 

-  _  ax1+bx2-\-cx^        _  ay1+by2  +  cyit 
a  +  h  +  c  y  "  a  +  b  +  c 

3.  If  two  parallel  forces,  P  and  Q,  act  in  the  same  direc- 
tion at  A  and  B,  (Fig.  14),  and  make  an  angle,  0,  with 
AB,  find  the  moment  of  each  about  the  point  of  applica- 
tion of  their  resultant. 

The  moment  of  P  with  respect  to  G  is 

P-  AG  sin  0  (Art.  46). 
But  from  (1)  of  Art.  45,  we  have 

P+Q__ABm 

Q      ~  AG> 

which  in  P  •  A  G  sin  0  gives 

for  the  moment  of  P  which  also  equals  the  moment  of  Q 


R'+— 

B 

R 

g/ 

V 

lw    c 

T 

EXAMPLES.  79 

4.  Two  parallel  forces,  acting  in  the  same  direction, 
have  their  magnitudes  5  and  13,  and  their  points  of  applica- 
tion, A  and  B,  6  feet  apart.  Find  the  magnitude  of  their 
resultant,  and  the  point  of  application,  G. 

Ans.  R  =  18,  AG  =  4J,  BG  =  If. 

5.  On  a  straight  rod,  AF,  there  are  suspended  5  weights 
of  5,  15,  7,  6,  and  9  pounds  respectively  at  the  points  A,  B, 
D,  F,  F;  AB  =  '6  feet,  BD  =  6  feet,  DE  =  5  feet, 
EF  =  4  feet.  Find  the  magnitude  of  the  resultant,  and 
the  distance  of  its  point  of  application,  G,  from  A. 

Ans.  R  =  42  pounds.  A  G  =  8f  feet. 

6.  A  heavy  uniform  beam,  AB,  rests 
in  a  vertical  plane,  with  one  end,  A,  on  a 
smooth  horizontal  plane  and  the  other 
end,  B,  against  a  smooth  vertical  wall ; 
the  end,  A,  is  prevented  from  sliding  by  a  n  2~ 
a  horizontal  string  of  given  length  fas- 
tened to  the  end  of  the  beam  and  to  the  wall ;  determine 
the  tension  of  the  string  and  the  pressures  against  the 
horizontal  plane  and  the  wall. 

Let  2a  =~the  length  of  the  beam,  and  let  W  be  its  weight, 
which  as  the  beam  is  uniform,  we  may  suppose  to  act  at  its 
middle  point,  G.  Let  R  be  the  vertical  pressure  of  the 
horizontal  plane  against  the  beam ;  and  R'  the  horizontal 
pressure  of  the  vertical  wall,  and  T  the  tension  of  the.  hor- 
izontal string,  AC ;  let  BAO  =  a,  a  known  angle,  since 
the  lengths  of  the  beam  and  the  string  are  given.  Then 
(Art.  61),  we  have 

for  horizontal  forces,  T  =  R' ; 

for  vertical  forces,       W  =  R; 

for  moments  about  A  (Art.  47),  2R'  a  sin  «  =  Wa  cos  «; 

W 
.-.     R'  =  T=^r  cot«. 
A 


80 


EXAMPLES. 


FJg.28 


7.  A  heavy  beam,  AB  =  a  +  b,  rests 
on  two  given  smooth  planes  which  are 
inclined  at  angles,  a  and  j3,  to  the 
horizon;  required  the  angle  6  which 
the  beam  makes  with  the  horizontal 
plane,  and  the  pressures  on  the 
planes. 

Let  a  and  b  be  the  segments,  AG  and  BG,  of  the  beam, 
made  by  its  centre  of  gravity,  G;  let  R  and  R'  be  the 
pressures  on  the  planes,  AC  and  BC,  the  lines  of  action  of 
which  are  perpendicular  to  the  planes  since  they  are  smooth, 
and  let  IF  be  the  weight  of  the  beam.     Then  we  have 

for  horizontal  forces,  R  sin  «  =  R'  sin  j3;  (1) 

for  vertical  forces,  R  cos  «  -f-  R!  cos  0  =  W ;       (2) 

for  moments  about  G,  Ra  cos  («  +  0)  =  R'b  cos  ((3—6).  (3) 

Dividing  (3)  by  (1),  we  have 

a  cot  a  —  a  tan  d  =  b  cot  fi  -\-  b  tan  6 ; 

a  cot  a  —  b  cot  /3 


therefore, 


tan  0  = 


a  +  b 


and  from  (1)  and  (2)  we  have 
W  sin  0 


72  = 


sin  («  +  j3) ' 


and  72'  = 


IF  sin  of 

sin  («  +  0)' 


Otherwise  thus:  since  the  beam  is  in  equilibrium  undei 
the  action  of  only  three  forces,  they  must  meet  in  a  point  0. 
(Art.  62),  and  therefore  we  obtain  immediately  from  the 
geometry  of  the  figure, 


sin  (3 


R 


W       sin  («+£) 


R  = 


W  sin  ft 
sin  (a  +  ay 


EXAMPLES. 


51 


and 


R' 


sin  a 


R'  = 


W  sin  cc 


)V~  sin  («  +  j3)'  sin  («  +  0) 

Also  since  the  angles,  GOA  and  GOB,  are  equal  to  a  and  3, 
respectively,  and  BGO  =  -  —  0,  we  have 

(a  +  b)  cot  BGO  =  a  cot  GOA.  —  b  cot  GOB; 
a  cot  «  —  b  cot  j3 


therefore,  tan  0  = 

_.  .,  «       tan  « 

Hence,  it  T  = jr, 

J        tan  j3 

position. 


«  +  b 
the  beam  will  rest  in  a  horizontal 


8.  A  heavy  uniform  beam,  AB,  rests  with 
one  end,  A,  against  a  smooth  vertical  wall, 
and  the  other  end,  B,  is  fastened  by  a  string, 
BC,  of  given  length  to  a  point,  C,  in  the 
wall ;  the  beam  and  the  string  are  in  a  vertical 
plane ;  it  is  required  to  determine  the  pressure 
against  the  wall,  the  tension  of  the  string,  and 
the  position  of  the  beam  and  the  string. 


Let 


AG  =  GB  =.  a,    AC  =  x,    BO  =  b, 


weight  of  beam  =  W,  tension  of  string  =  T,  pressure  of 
wall  =  R, 

BAE  =  e,     BOA  =  0. 
Then  we  have 

for  horizontal  forces,  R  =  T  sin  0;  (1) 

for  vertical  forces,        W  =  T  cos  0 ;  (2) 

for  moments  about  A,  Wa  sin  6  =  T-  AD  =  Tx  sin  0;  (3) 

.  •.    a  sin  6  =  x  tan  (p ;  (4) 


82 

EXAMPLES. 

and  by  the  geometry 

of  the  1 

igure 

b  _ 

2a 

sin  0 
"  sin^' 

X 

2a  ~ 

_  sin  (d—(f)) 
sin  <f> 

Solving  (4),  (5), 

and 

(6),  we 

get 

..[> 

3      J» 

(5) 
(6) 


2  f>  —  4tfH* 
008  *  =  3  L- 3 — J  ; 


sin 


1  fl6a2  —  ^ 
^  =  2^L 3-~J  ; 


from    which    R    and    T  become  known.     (Price's  Anal. 
Mech's.,  Vol.  I,  p.  69). 

To  determine  all  the  unknown  quantities  many  problems  in  Statics 
require  equations  to  be  formed  by  geometric  relations  as  well  as  static 
relations.  Thus  (1),  (2),  (3)  are  static  equations,  and  (5)  is  a  geometric 
equation. 

9.  A  uniform  heavy  beam,  AB  =  2a, 
rests  with  one  end,  A,  against  the  inter- 
nal surface  of  a  smooth  hemispherical 
bowl,  radius  =  r,  while  it  is  supported 
at  some  point  in  its  length  by  the  edge 
of  the  bowl ;  find  the  position  of  equili-  7i^3o 

briu  m. 

The  beam  is  kept  in  equilibrium  by  three  forces,  viz.,  the 
reaction,  R,  at  A  perpendicular  to  the  surface  of  contact, 
(Art.  42)  and  therefore  perpendicular  to  the  bowl,  the 
reaction,  R',  at  C  which,  for  the  same  reason,  is  perpen* 
dicular  to  the  beam,  and  the  weight  W  acting  at  G. 


EXAMPLES.  83 

Let  6  =  the  inclination  of  the  beam  to  the  horizon 
=  <ACD.  The  solution  will  be  most  readily  effected  by 
resolving  the  forces  along  the  beam  and  taking  moments 
about  C,  by  which  we  shall  obtain  equations  free  from  the 
unknown  reaction,  R\     Then  we  have 

for  forces  along  AB,  R  cos  0  =  W  sin  0,  (I) 

for  moments  about  0, 

R  •  2r  cos  0  sin  0  =  W  (2r  cos  0  —  a)  cos  0.         (2) 

From  (1)  we  have 

R  =  W  tan  0, 

which  in  (2)  gives,  after  reducing, 

2r  sin2  B  —  2r  cos2  6  +  a  cos  6  =  0, 

or,  4r  cos2  6  —  a  cos  6  —  2r  =  0,  (3) 


cos  (9  = ! 

Sr 


Otherwise  thus:  since  the  beam  is  in  equilibrium  under 
the  action  of  only  three  forces,  they  must  meet  in  a  point 
0  (Art.  62).  Draw  the  three  forces  AG,  00,  GO,  which 
keep  the  beam  in  equilibrium.  Let  the  line,  OG,  meet  the 
semicircle,  DAC,  in  the  point,  Q.  Then  AQ  is  a  horizontal 
line.     Also 

<QAG  =  <DCA  =  0, 
therefore  <OAQ  ==  20. 

Hence  AQ  =  AO  cos  20, 

and  also  AQ  =  AG  cos  0 ; 


S4 


EXAMPLES. 


Fig.3l 


therefore  2r  cos  20  =  a  cos  0, 

or  4r  cos2  0  —  a  cos  0  —  2r  =  0, 

which  is  the  same  as  (3)  obtained  by  the  other  method. 
The  student  may  prove    that    the    reaction,    R\  at  0 

2r 

10.  Find  the  position  of  equilibrium  of 
a  uniform  heavy  beam,  one  end  of  which 
rests  against  a  smooth  vertical  plane,  and 
the  other  against  the  internal  surface  of  a 
smooth  spherical  bowl. 

The  beam  is  in  equilibrium  under  the 
action   of  three   forces,    the   weight,   W, 
acting  at  G-,  the  reaction,  R,  at  A,  perpen- 
dicular to  the  surface  and  hence  passing  through  the  centre, 
C,  and  the  reaction,  R'9  of  the  vertical  plane  perpendicular 
to  itself  and  hence  horizontal. 

Let  the  length  of  the  beam,  AB,  =  2a,  r  =  the  radius 
of  the  sphere,  d  =  CD,  the  distance  of  the  centre  of  the 
sphere  from  the  vertical  wall,  W  =  the  weight  of  the  beam  ; 
and  let  0  =  the  required  inclination  of  the  beam  to  the 
horizon,  and  0  =  the  inclination  of  the  radius  AC  to  the 
horizon.    Then  we  have 

for  vertical  forces,  R  sin  0  =  W;  (1) 

for  moments  about  B,  R-  2a  sin  (</>— 0)  =  W>  a  cos  6 ;  (2) 

Dividing  (2)  by  (1)  we  have 

2  sin  (0  —  6) 


or 


sin  </> 
tan  0  =  2  tan  0 


cos  0, 


(3) 


CENTRE    OF    PARALLEL    FORCES.  85 

Then  we  have,  from  the  geometry  of  the  figure,  the 
horizontal  distance  from  A  to  the  wall  =  the  horizontal 
projection  of  AC  +  CD,  that  is, 

2a  cos  0  =  r  cos  0  -f  d.  (4) 

From  (3)  and  (4)  a  value  of  6  can  be  obtained,  and  hence 
the  position  of  equilibrium. 

Otherwise  thus:  since  the  beam  is  in  equilibrium  under 
the  action  of  only  three  forces  they  must  meet  in  a  point,  0. 
Geometry  then  gives  us 

2  cot  OGB  =  cot  AOG  -  cot  GOB  =  cot  AOG, 

or  2  tan  d  =  tan  </>, 

which  is  the  same  as  (3). 

63.  Centre  of  Parallel  Forces  in  Different  Planes. 

~~To  find  the  magnitude,  direction,  and  point  of 
application  of  the  resultant  of  any  number  of 
parallel  forces  acting  on  a  rigid  body. 

The  theorem  of  Art.  59  is  evidently  true  also  in  the  case 
in  which  neither  the  parallel  forces  nor  their  fixed  points  of 
application  lie  in  the  same  plane,  hence,  calling  i  the  third 
co-ordinate  of  the  point  of  application  of  the  resultant,  we 
have  for  the  distance  of  the  centre  of  parallel  forces  from 
the  planes  yz,  zx,  and  xy, 

.  _  ZPx      -  _  !J>y  _  "LPz 

x~  ?>p>    y  -  zp>    s-  XP* 

Hence  (Art.  59,  Sch.)  the  equations  for  determining  the 
position  of  the  centre  of  parallel  forces  show  that  the  sum 
of  the  moments  of  the  parallel  forces  with  respect  to  any 
plane  is  equal  to  the  moment  of  their  resultant. 


yy 


L|R 


Fig.32 


86     EQUILIBRIUM  OF  PARALLEL  FORCES  IN  SPACES. 

64.  Conditions  of  Equilibrium  of  a  System  of 
Parallel  Forces  Acting  upon  a  Rigid  Body  in 
Space. — Let  P19  P2,  Pz,  etc.,  denote  the  forces,  and  let 
thern  be  referred  to  three  rectangular  axes, 
OX,  OY,  0Z\  the  last  parallel  to  the 
forces  ;  let  {xx,  yu  zt),  (x2,  y2,  z2),  etc., 
be  the  points  of  application  of  the  forces, 
Pi9  P2,  etc.  Let  the  direction  of  Px 
meet  the  plane,  xy,  at  Mt. 

Draw  MxNt  perpendicular  to  the  axis 
of  x  meeting  it  at  Nx.  Apply  at  0,  and  also  at  Nu  two 
opposing  forces  each  equal  and  parallel  to  Pv  Then  the 
force  Pt  at  Mt  is  replaced  by 

(1)  Px  at  0  along  OZ, 

(2)  a  couple  formed  of  P^  at  Mt  and  Pt  at  Nx  ; 

(3)  a  couple  formed  of  Pt  at  Nt  and  Px  at  0. 

The  moment  of  the  first  couple  is  P-^y^  and  this  couple 
may  be  transferred  to  the  plane  yz,  which  is  parallel  to  its 
original  plane,  without  altering  its  effect  (Art.  52).  The 
moment  of  the  second  couple  is  Ptxt,  and  the  couple  is  in 
the  plane  xz. 

Replacing  each  force  in  this  manner,  the  whole  system 
will  be  equivalent  to  a  force 

pi  +  p2  +  Fs  +  etc-;  or  2P  at  0  along  OZ. 
together  with  the  couple 

p\Vi  +ps#2  +PzVz  +etc,  or  ZPy,  in  the  plane  yz, 

and  the  couple 

Ptxx  +  P2x2  +Pzx3  +etc,  or  ZPx  in  the  plane  xz. 

The  first  couple  tends  to  turn  the  body  from  the  axis  of  y 
fco  that  of  z  round  the  axis  of  x,  and  the  second  couple 


EQUILIBRIUM  OF  PARALLEL   FORCES  IN  SPACE.      8? 

tends  to  turn  the  body  from  the  axis  of  x  to  that  of  z 
round  the  axis  of  y.  It  is  customary  to  consider  those 
couples  as  positive  which  tend  to  turn  the  body  in  the 
direction  indicated  by  the  natural  order  of  the  letters,  i.  e.y 
positive  from  x  to  y,  round  the  z-axis ;  from  y  to  z  round 
the  ar-axis ;  and  from  z  to  x  round  the  ?/-axis  ;  and 
negative  in  the  contrary  direction. 

Hence  the  moment  of  the  first  couple  is  -\-ZPy,  aud 
therefore  OX  is  its  axis  (Art.  50) ;  and  the  moment  of 
the  second  couple  is  —  ZPx,  and  therefore  OY'  is  its  axis. 
The  resultant  of  these  two  couples  is  a  single  couple  whose 
axis  is  found  (Art.  56)  by  drawing  OL  (in  the  positive 
direction  of  the  axis  of  x)  =  ZPy,  and  OM  (in  the  nega- 
tive direction  of  the  axis  of  y)  =  ZPx,  and  completing  the 
parallelogram  OLGM.  If  OG,  the  diagonal,  is  denoted  by 
G,  we  have 

G  =  V(zPzy  +  Wy)\ 

and  R  =  SP; 

R  being  the  resultant  force. 

Now  since  this  single  force,  R,  and  this  single  couple,  G, 
cannot  produce  equilibrium  (Art.  54,  Cor.),  we  must  have 
R  =  0,  and  G  =  0,  and  G  cannot  be  =  0  unless  ZPx  =  0 
and  I*Py  =  0;  the  conditions  therefore  of  equilibrium  are 

R  =  0, 

IPx  =  0,     ZPy  =  0. 

Hence,  the  conditions  of  equilibrium  of  parallel  forces  in 
space  are : 

TJie  sum  of  the  forces  must  =  0. 

The  sum  of  the  moments  of  the  forces  with  respect  to 
every  plane  parallel  to  them  must  =  0. 


z 

o 

V 

A, 

(n,  , 

/ 

( 

A 

88  EQUILIBRIUM  OF  FORCES. 

65.  Conditions  of  Equilibrium  of  a  System  of 
Forces  acting  in  any  Direction  on  a  Rigid  Body  in 
Space. — Let  Pl9  P2,  P3,  etc.,  denote  the  forces,  and  let 
them  be  referred  to  three  rectangular  axes,  OX,  OY,  0Z\ 
let  (xx,  yx,  zx),  (x2,  y2,  z2),  etc.,  be  the  points  of  applica- 
tion of  Px,  P2,  etc. 

Let  Ax  be  the  point  of  application  of 
Px;  resolve  Px  into  components  XX9 
Yt,  Zx,  parallel  to  the  co-ordinate  axes. 
Let  the  direction  of  Zx  meet  the  plane 
xy  at  M19  and  draw  MXNX  perpendicu- 
lar to  OX.  Apply  at  Xx  and  also  at  0 
two  opposing  forces  each  equal  and  par- 
allel to  Zx.  Hence  Zx  at  Ax  or  Mx  is  equivalent  to  Zx  at 
0,  and  two  couples  of  which  the  former  has  its  moment  = 
Zx  x  NXMX  =  Z1y1,  and  may  be  supposed  to  act  in  the 
plane  yz,  and  the  latter  has  its  moment  =  Zx  x  0NX  = 
—  Zxxx  and  acts  in  the  plane  zx. 

Hence  Zx  is  replaced  by  Z,  at  0,  a  couple  Z1y1  in  the 
plane  yz,  and  a  couple  —  Zxxx  (Art.  64)  in  the  plane  zx. 
Similarly  Xx  may  be  replaced  by  Xx  at  0,  a  couple  Xxzx 
in  the  plane  zx,  and  a  couple  —  -3Ti#i  m  the  plane  xy. 
And  T^  may  be  replaced  by  Y1  at  0,  a  couple  F^  in  the 
plane  xy,  and  a  couple  —  J^i  in  the  plane  yz.  Therefore 
the  force  Px  may  be  replaced  by  Xx,  Y19  Zx,  acting  at  0, 
and  three  couples,  of  which  the  moments  are,  (Art.  56), 

Zxyx  —  Yxzt  in  the  plane  yz,  around  the  axis  of  x, 
Xxzx  —  Zxxx  in  the  plane  zx,  around  the  axis  of  y, 
Y1x1  —  Xxyx  in  the  plane  xy,  around  the  axis  of  z. 

By  a  similar  resolution  of  all  the  forces  we  shall  have 
them  replaced  by  the  forces 

SJ,     2  7,     ZZ, 

acting  at  0  along  the  axes,  and  the  couples 


EQUILIBRIUM    OF  FORCES.  89 

Z  (Zy  —  Yz)  =  L,  suppose,  in  the  plane  yz, 
Z  (Xz  —  Zx)  —  M,  suppose,  in  the  plane  zx, 
Z  (Yx  —  Xy)  =  N,  suppose,  in  the  plane  xy. 

Let  R  be  the  resultant  of  the  forces  which  act  at  0;  a> 
£,  c,  the  angles  its  direction  makes  with  the  axes ;  then 
(Art.  38), 

R?  =  (SX)a  +  (ZYf  +  (^Z)% 

ZX  .       ZY  ZZ 

cos  a  =  — q-,    cos  o  =  —&-,    cos  c  =  -5-. 

XL  it  K 

Let  67  be  the  moment  of  the  couple  which  is  the  result- 
«nt  of  the  three  couples,  L,  M,  N;  X,  /j,,  v,  the  angles  its 
axis  makes  with  the  co-ordinate  axes  ;  then  (Art.  56,  Sch.), 

672  =  D  +  M*  +  IP, 

L  M  N 

cos  A  =  — ,    cos  \i  =  -^j     cos  v  =  — . 

6r  Or  6r 

Therefore  any  system  of  forces  acting  in  any  direction  on 
a  rigid  body  in  space  may  always  be  reduced  to  a  single 
force,  R.  and  a  single  couple,  67,  and  cannot  therefore  pro- 
duce equilibrium  (Art.  54,  Cor.).  Hence  for  equilibrium 
we  must  have  R  =  0  and  G  =  0  ;  therefore 

(sj)2  +  (217  +  (ZZf  =  0, 
and  Z2  +  3P  +  N*  =  0. 

These  lead  to  the  six  conditions, 

ZX=0,     ZY  =  0,     ZZ=0, 
Z  (Zy  -  Fs)  =  0,  .2  ( JTs  -  Zs)  =  0? 

E(r^-x^)  =  0. 


90  EXAMPLES. 


EXAMPLES 


1.  If  the  weights,  1,  2,  3,  4,  5  lbs.,  act  perpendicularly 
to  a  straight  line  at  the  respective  distances  of  1,  2,  3,  4, 
5  feet  from  one  extremity,  find  the  resultant,  and  the  dis- 
tance of  its  point  of  application  from  the  first  extremity. 

Am.  R  =  15  lbs.,  x  =  3f  feet. 

2.  Four  weights  of  4,  —7,  8,  —3  lbs.,  act  perpendicularly 
to  a  straight  line  at  the  points  A,  B,  C,  D,  so  that  AB  = 
5  feet,  BC  =  4  feet,  CD  =  2  feet ;  find  the  resultant  and 
its  point  of  application,  G. 

Am.  R  =  2  lbs.,  AG  =  2  feet. 

3.  Two  parallel  forces  of  23  and  42  lbs.,  act  at  the  points 
A  and  B,  14  inches  apart;  find  GB  to  three  places  of 
decimals.  Ans.  4  954  ins. 

4.  Two  weights  of  3  cwts.  2  qrs.  15  lbs.,  and  1  cwt.  3  qrs. 
25  lbs.  are  supported  at  the  points  A  and  B  of  a  straight 
line,  the  length  AB  =  3  feet  7  inches ;  find  AG  to  three 
places  of  decimals  of  feet.  Ans.  1.268  ft. 

5.  A  bar  of  iron  15  inches  long,  weighing  12  lbs.,  and  of 
uniform  thickness,  has  a  weight  of  10  lbs.  suspended  from 
one  extremity  ;  at  what  point  must  the  bar  be  supported 
that  it  may  just  balance. 

The  weight  of  the  bar  acts  at  its  centre. 

Ans.  4^j  in.  from  the  weight. 

6.  A  bar  of  uniform  thickness  weighs  10  lbs.,  and  is 
5  feet  long  ;  weights  of  9  lbs.  and  5  lbs.  are  suspended  from 
its  extremities  ;  on  what  point  will  it  balance  ? 

Ans.  5  in.  from  the  centre  of  the  bar. 

7.  A  beam  30  feet  long  balances  itself  on  a  point  at  one- 
third  of  its  length  from  the  thicker  end  ;  but  when  a  weight 
of  10  lbs.  is  suspended  from  the  smaller  end,  the  proj*  must 


EXAMPLES.  91 

be  moved  two  feet  towards  it,  in  order  to  maintain  the 
equilibrium.     Find  the  weight  of  the  beam.  Am*  90  lbs. 

8.  A  uniform  bar,  4  feet  long,  weighs  10  lbs.,  and  weights 
of  30  lbs.  and  40  lbs.  are  appended  to  its  two  extremities  ; 
where  must  the  fulcrum*  be  placed  to  produce  equilibrium  ? 

Ans,  3  in.  from  the  centre  of  the  bar. 

9.  A  bar  of  iron,  of  uniform  thickness,  10  ft.  long,  and 
weighing  1J  cwt,,  is  supported  at  its  extremities  in  a  hori- 
zontal position,  and  carries  a  weight  of  4  cwt.  suspended 
from  a  point  distant  3  ft.  from  one  extremity.  Find  the 
pressures  on  the  points  of  support. 

Ans.  3.55  cwt,,  and  1.95  cwt. 

10.  A  bar,  each  foot  in  length  of  which  weighs  7  lbs., 
rests  upon  a  fulcrum  distant  3  feet  from  one  extremity ; 
what  must  be  its  length,  that  a  weight  of  71^  lbs.  sus- 
pended from  that  extremity  may  just  be  balanced  by 
20  lbs.  suspended  from  the  other  ?  Ans.  9  ft. 

11.  Five  equal  parallel  forces  act  at  5  angles  of  a  regular 
hexagon,  whose  diagonal  is  a  ;  find  the  point  of  application 
of  their  resultant. 

Ans.  On  the  diagonal  passing  through  the  sixth  angle,  at 
a  distance  from  it  of  f  a. 

12.  A  body,  P,  suspended  from  one  end  of  a  lever  with- 
out weight,  is  balanced  by  a  weight  of  1  lb.  at  the  other 
end  of  the  lever ;  and  when  the  fulcrum  is  removed 
through  half  the  length  of  the  lever  it  requires  10  lbs.  to 
balance  P  ;  find  the  weight  of  P.      Ans.  5  lbs.  or  2  lbs. 

13.  A  carriage  wheel,  whose  weight  is  W  and  radius  r, 
rests  upon  a  level  road;  show  that  the  force,  F9  necessary 
to  draw  the  wheel  over  an  obstacle,  of  height  h,  is 

r  —  h 

*  The  support  on  which  it  rests. 


92  EXAMPLES. 

14.  A  beam  of  uniform  thickness,  5  feet  long,  weighing 
10  lbs.,  is  supported  on  two  props  at  the  ends  of  the  beam; 
find  where  a  weight  of  30  lbs.  must  be  placed,  so  that  the 
pressures  on  the  two  props  may  be  15  lbs.  and  25  lbs. 

Aiis.  10  ins.  from  the  centre. 

15.  Forces  of  3,  4,  5,  6  lbs.  act  at  distances  of  3  ins., 
4  ins.,  5  ins.  6  ins.,  from  the  end  of  a  rod  ;  at  what  distance 
from  the  same  end  does  the  resultant  act? 

Ans.  4|  inches. 

16.  Four  vertical  forces  of  4,  6,  7,  9  lbs.  act  at  the  four, 
corners  of  a  square ;  find  the  point  of  application  of  the 
resultant.     Ans.  T53  of  middle  line  from  one  of  the  sides. 

17.  A  flat  board  12  ins.  square  is  suspended  in  a  hori- 
zontal position  by  strings  attached  to  its  four  corners,  A, 
B,  C,  D,  and  a  weight  equal  to  the  weight  of  the  board  is 
laid  upon  it  at  a  point  3  ins.  distant  from  the  side  AB  and 
4  ins.  from  AD  ;  find  the  relative  tensions  in  the  four 
strings.  Ans.  As  }  :  J  :  J  :  -{%. 

18.  A  rod,  AB,  moves  freely  about  the  end,  B,  as  on  a 

hinge.     Its  weight,  W,  acts  at  its  middle  point,  and  it  is 

kept  horizontal  by  a  string,  AC,  that  makes  an  angle  of  45° 

with  it.     Find  the  tension  in  the  string.  .  W 

°  Ans.  — -• 

V2 

19.  A  rod  10  inches  long  can  turn  freely  about  one  of 
its  ends  ;  a  weight  of  4  lbs.  is  slung  to  a  point  3  ins.  from 
this  end,  and  the  rod  is  held  by  a  string  attached  to  its  free 
end  and  inclined  to  it  at  an  angle  of  120°  ;  find  the 
tension  in  the  string  when  the  rod  is  horizontal. 

Ans.  |  V3  lbs. 

20.  Two  forces  of  3  lbs.  and  4  lbs.  act  at  the  extremities 
of  a  straight  lever  12  ins.  long,  and  inclined  to  it  at  angles 
of  120°  and  135°  respectively  ;  find  the  position  of  the 
fulcrum.  j^nSm   (8  —  3  a/6)  x  9.6  ins.  from  one  end. 


EXAMPLES.  93 

9,1.  Find  the  true  weight  of  a  body  which  is  found  to 
weigh  8  ozs.  and  9  ozs.  when  placed  in  each  of  the  scale- 
pans  of  a  false  balance.  £m%  G  ^/%  ozs> 

22.  A  beam  3  ft.  long,  the  weight  of  which  is  10  lbs., 
and  acts  at  its  middle  point,  rests  on  a  rail,  with  4  lbs.  hang- 
ing from  one  end  and  13  lbs.  from  the  other  ;  find  the  point 
at  which  the  beam  is  supported  ;  and  if  the  weights  at  the 
two  ends  change  places,  what  weight  must  be  added  to  the 
lighter  to  preserve  equilibrium  ? 

Ans.  12  ins.  from  one  end  ;  27  lbs. 

23.  Two  forces  of  4  lbs.  and  8  lbs.  act  at  the  ends  of  a 
bar  18  ins.  long  and  make  angles  of  120°  and  90°  with  it; 
find  the  point  in  the  bar  at  which  the  resultant  acts. 

A?is.  -ff  (4  —  V3)  ins.  from  the  4  lbs.  end. 

24.  A  weight  of  24  lbs.  is  suspended  by  two  flexible 
strings,  one  of  which  is  horizontal,  and  the  other  is  inclined 
at  an  angle  of  30°  to  the  vertical.  What  is  the  tension  in 
each  string  ?  AnSm  8  ^3  lbs> .  16  ^3  lbg< 

25.  A  pole  12  ft.  long,  weighing  25  lbs.,  rests  with  one 
end  against  the  foot  of  a  wall,  and  from  a  point  2  ft.  from 
the  other  end  a  cord  runs  horizontally  to  a  point  in  the 
wall  8  ft.  from  the  ground  ;  find  the  tension  of  the  cord  and 
the  pressure  of  the  lower  end  of  the  pole. 

Ans.  11.25  lbs.;  27.4  lbs. 

26.  A  body  weighing  6  lbs.  is  placed  on  a  smooth  plane 
which  is  inclined  at  30°  to  the  horizon  ;  find  the  two  direc- 
tions in  which  a  force  equal  to  the  body  may  act  to  produce 
equilibrium.  Also  find  what  is  the  pressure  on  the  plane 
in  each  case. 

Ans.  A  force  at  60°  with  the  plane,  or  vertically  upwards  ; 

R  =  6  a/3,  or  0. 

27.  A  rod,  AB,  5  ft.  long,  without  weight,  is  hung  from 
a  point,  0,  by  two  strings  which   are  attached  to  its  ends 


94 


EXAMPLES. 


and  to  the  point ;  the  string,  AC,  is  3  ft.,  and  BC  is  4  ft.  in 
length,  and  a  weight  of  2  lbs.  is  hung  from  A,  and  a  weight 
of  3  lbs.  from  B  ;  find  the  tensions  of  the  strings. 

Ans.  V5  1bs.;  2  V5  lbs. 

28.  Find  the  height  of  a  cylinder,  which  can  just  rest  on 
an  inclined  plane,  the  angle  of  which  is  60°,  the  diameter 
of  the  cylinder  being  6  ins.  and  its  weight  acting  at  the 
middle  point  of  its  axis.  Ans.  3.46  ins. 

29.  Two  equal  weights,  P,  Q,  are  connected  by  a  string 
which  passes  over  two  smooth  pegs,  A,  B,  situated  in  a 
horizontal  line,  and  supports  a  weight,  W,  which  hangs 
from  a  smooth  ring  through  which  the  string  passes ;  find 
the  position  of  equilibrium. 

Ans.  The  depth  of  the  ring  below  the  line 

W 


AB  = 


2  V4P2  -  W* 


AB. 


30.  The  resultant  of  two  forces,  P,  Q,  acting  at  an  angle, 
0,  is  =  (2m  +  1)  \/P2  +  Q2;  when  they  act  at  an  angle, 

%  —  B,  it  is  =  (2m  —  1)  a/P2  +  Qz;   show  that  tan  d  = 

m  —  1 

m  +T 

31.  A  uniform  heavy  beam,  AB  =  2«, 
rests  on  a  smooth  peg,  P,  and  against  a 
smooth  vertical  wall,  AD ;  the  horizontal 
distance  of  the  peg  from  the  wall  being 
h ;  find  the  inclination,  0,  of  the  beam  to 
the  vertical,  and  the  pressures,  R  and  S,  on  the  wall  and  peg. 

32.  Two  equal  smooth  cylinders  rest  in  contact  on  two 
smooth  planes  inclined  at  angles,  a  and  |3,  to  the  horizon; 


EXAMPLES.  95 

find  the  inclination,  0,  to  the  horizon  of  the  line  joining 
their  centres.  Ans.  tan  0  =  \  (cot  a  —  cot  (3). 

33.  A  beam,  5  ft.  long,  weighing  5  lbs.,  rests  on  a  ver- 
tical prop,  CD  =  2£  ft. ;  the  lower  end,  A,  is  on  a  hori- 
zontal plane,  and  is  prevented  from  sliding  by  a  string, 
AD  =  3£  ft. ;  find  the  tension  of  the  string. 

Ans.   T=%$  lbs. 

34.  A  uniform  beam,  AB,  is  placed  with  one  end,  A, 
inside  a  smooth  hemispherical  bowl,  with  a  point,  P,  rest- 
ing on  the  edge  of  the  bowl.  If  AB  =  3  times  the  radius 
R9  find  AP.  Ans.  AP  =  1.838  R. 

35.  A  body,  weight  W,  is  suspended  by  a  cord,  length  ?, 

from  the  point  A,  in  a  horizontal  plane,  and  is  thrust  out 

of  its  vertical  position  by  a  rod  without  weight,  acting  at 

another    point,   B,   in    the    horizontal    plane,    such    that 

AB  =  d,  and  making  the  angle,  0,  with  the  plane ;  find 

the  tension,  T,  of  the  cord.  =       I  ^ 

a 
y 

36.  Two  heavy  uniform  bars,  AB  and 

CD,  movable  in  a  vertical  plane  about 
their  extremities,  A,  D,  which  rest  on  a 
horizontal  plane  and  are  prevented  from 
sliding  on  it;  find  their  position  of 
equilibrium  when  leaning  against  each 
other. 

Let  the  bars  rest  against  each  other  at  B,  and  let 
AD  =  a,  AB  =  h,  CD  =  c,  BD  =  x,  W  and  Wt  =  the 
weights  of  AB  and  CD,  respectively  acting  at  their  middle 
points ;  then  we  have 

2x*W  (a2  -f  5*—  x*)  =  cW1  (a2  -f-  a?  —  &)  (J2  +  x*  —  a2), 

which  is  an  equation  of  the  fifth  degree,  and  hence  always 
has  one  real  root,  the  value  of  which  may  be  determined 
when  numbers  are  put  for  a>  b,  and  c. 


96 


EXAMPLES. 


37.  A  parabolic  curve  is 
placed  in  a  vertical  plane  with 

its  axis  vertical  and  vertex 
downwards,  and  inside  of  it, 
and  against  a  peg  in  the  focus, 
and  against  the  concave  arc,  a 
smooth  uniform  and  heavy 
beam  rests ;  required  the  posi- 
tion of  equilibrium. 

Let  PB  be  the  beam,  of 
length  7,  and  of  weight  W, 
resting  on  the  peg  at  the  focus, 
F ;  let  AF  =  p  and  AFP  =  0. 


38.  Find  the  form  of  the  curve  in  a  vertical  plane  such 
that  a  heavy  bar  resting  on  its  concave  side  and  on  a  peg  at 
a  given  point,  say  the  origin,  may  be  at  rest  in  all 
positions. 

Am.  r  =  Jl  +  h  sec  0,  in  which  I  =  the  length  of  the 
bar,  k  an  arbitrary  constant,  and  6  the  inclination  of  the 
bar  to  the  vertical.  It  is  the  equation  of  the  conchoid  of 
Nicomedes. 

39.  A  rod  whose  centre  of  gravity  is  not  its  middle  point 
is  hung  from  a  smooth  peg  by  means  of  a  string  attached 
to  its  extremities  ;  find  the  position  of  equilibrium. 

A?is.  There  are  two  positions  in  which  the  rod  hangs 
vertically,  and  there  is  a  third  thus  defined : — Let  F  be  the 
extremity  of  the  rod  remote  from  the  centre  of  gravity,  k 
the  distance  of  the  centre  of  gravity  from  the  middle  point 
of  the  rod,  2a  the  length  of  the  string,  and  2c  the  length  of 
the  rod ;  then  measure  on  the  string  a  length  FP  from  F 

equal  to  a  ( L  -\ — I,  and  place  the  point  P  over  the  peg. 

This  will  define  a  third  position  of  equilibrium. 


EXAMPLES.  97 

40.  A  smooth  hemisphere  is  fixed  on  a  horizontal  plane, 
with  its  convex  side  turned  upwards  and  its  base  lying  in 
the  plane.  A  heavy  uniform  beam,  AB,  rests  against  the 
hemisphere,  its  extremity  A  being  just  out  of  contact  with 
the  horizontal  plane.  Supposing  that  A  is  attached  to  a 
rope  which,  passing  over  a  smooth  pulley  placed  vertically 
over  the  centre  of  the  hemisphere,  sustains  a  weight,  find 
the  position  of  equilibrium  of  the  beam,  and  the  requisite 
magnitude  of  the  suspended  weight. 

Ana.  Let  W  be  the  weight  of  the  beam,  2a  its  length,  P 
the  suspended  weight,  r  the  radius  of  the  hemisphere,  h 
the  height  of  the  pulley  above  the  plane,  0  and  <j>  the 
inclinations  of  the  beam  and  rope  to  the  horizon ;  then  the 
position  of  equilibrium  is  defined  by  the  equations. 

r  cosec  0  =  U  cot  </>,  (1) 

r  cosec2  0  =  a  (tan  <f>  -f  cot  0),  (2) 

which  give  the  single  equation  for  0, 

r  (r  —  a  sin  0  cos  0)  =  ah  sin3  0.  (3) 

sin  0 


Also  P  =  W 


cos  (0  —  6) 


^ a  sin*  OVr2  +  A2  sin2  0  ,AX 

=  w ^ —     W 

y 

41.  If,  in  the  last  example,  the  position  and  magnitude 
of  the  beam  be  given,  find  the  locus  of  the  pulley. 

Ans.  A  right  line  joining  A  to  the  point  of  intersection 

of  the  reaction  of  the  hemisphere  and  W. 
y 

42.  If,  in  the  same  example,  the  extremity,  A,  of  the 
beam  rest  against  the  plane,  state  how  the  nature  of  the 
problem  is  modified,  and  find  the  position  of  equilibrium. 

Ans.  The  suspended  weight  must  be  given,  instead  of 
being  a  result  of  calculation.     Equation  (1)  still  holds,  but 


98  EXAMPLES. 

not  (2)  ;  and  the  position  of  equilibrium  is  denned  by   the 
equation 

Ph2  cos3  <j>  =  War  sin3  «/>. 

43.  If  the  fixed  hemisphere  be  replaced  by  a  fixed  sphere 
or  cylinder  resting  on  the  plane,  and  the  extremity  of  the 
beam  rest  on  the  ground,  find  the  position  of  equilibrium. 

Am.  If  h  denote  the  vertical  height  of  the  pulley  above 
the  point  of  contact  of  the  sphere  or  cylinder  with  the 
plane,  we  have 

r  cot  -  =  h  cot  (p, 

Pr  (1  +  cot  -  cot  6)  cos  </>  =  Wa  cos  0. 

44.  One  end,  A,  of  a  heavy  uniform  beam  rests  against  a 
smooth  horizontal  plane,  and  the  other  end,  B,  rests  against 
a  smooth  inclined  plane  ;  a  rope  attached  to  B  passes  over 
a  smooth  pulley  situated  in  the  inclined  plane,  and  sustains 
a  given  weight;  find  the  position  of  equilibrium. 

Let  6  be  the  inclination  of  the  beam  to  the  horizon,  «  the 
inclination  of  the  inclined  plane,  W  the  weight  of  the  beam, 
and  P  the  suspended  weight  ;  then  the  position  of  equili- 
brium is  defined  by  the  equation 

cos  6  ( W  sin  a  —  2P)  =  0.  (1) 

Hence  we  draw  two  conclusions: — 

(a)  If  the  given  quantities  satisfy  the  equation  IF  sin  « 
—  2P  =  0,  the  beam  will  rest  in  all  positions. 

(b)  There  is  one  position  of  equilibrium,  namely,  that  in 
which  the  beam  is  vertical. 

This  position  requires  that  both  planes  be  conceived  as 
prolonged  through  their  line  of  intersection. 

45.  A  uniform  beam,  AB,  movable  in  a  vertical  plane 
about  a  smooth  horizontal  axis  fixed  at  one  extremity,  A,  is 


EXAMPLES.  99 

attached  by  means  of  a  rope  BC,  whose  weight  is  negligible, 
to  a  fixed  point  C  in  the  horizontal  line  through  A,  such 
that  AB  =  AC ;  find  the  pressure  on  the  axis. 

Ans.    If  0  =  <CAB,   W  =  weight   of    beam,    the    re- 
action is 


/ft  ft 

4  sin2-  +  sec2-' 


CHAPTER  IV. 

CENTRE  OF  GRAVITY*  (CENTRE  OF  MASS). 

66.  Centre  of  Gf-ravity. — Gravity  is  the  name  given  tc 
the  force  of  attraction  which  the  earth  exerts  on  all  bodies ; 
the  effects  of  this  force  are  twofold,  (1)  statical  in  virtue  of 
which  all  bodies  exert  pressure,  and  (2)  kinetical  in  virtue 
of  which  bodies  if  unsupported,  will  fall  to  the  ground 
(Art.  15).  The  force  of  gravity  varies  slightly  from  place 
to  place  on  the  earth's  surface  (Art.  23) ;  but  at  each  place 
it  is  a  force  exerted  upon  every  body  and  upon  every 
particle  of  the  body  in  directions  that  are  normal  to  the 
earth's  surface,  and  which  therefore  converge  towards  the 
earth's  centre ;  but  as  this  centre  is  very  distant  compared 
with  the  distance  between  the  particles  of  any  body  of 
ordinary  magnitude,  the  convergence  is  zo  small  that  the 
lines  in  which  the  force  of  gravity  acts  are  sensibly  parallel. 

The  centre  of  gravity  of  a  body  is  the  point  of  application 
of  the  resultant  of  all  the  forces  of  gravity  which  act  upon 
every  particle  of  the  body ;  and  since  these  forces  are 
practically  parallel,  the  problem  of  finding  its  position  may 
be  treated  in  the  same  way  as  that  of  finding  the  centre  of  a 
system  of  parallel  forces  (Arts.  45,  59,  63).  The  centre  of 
gravity  may  also  be  defined  as  the  point  at  which  the  whole 
weight  of  a  body  acts.  If  the  body  be  supported  at  this 
point  it  will  rest  in  any  position  whatever. 

The  weight  of  a  body  is  the  resultant  of  all  the  forces  of 
gravity  which  act  upon  every  particle  of  it,  and  is  equal  in 
magnitude  and  directly  opposite  to  the  force  which  will  just 
support   the   body.      Since   the   centre   of  gravity  is   here 

*  Called  also  Centre  of  Mass  and  Centre  of  Inertia ;  and  the  term  Centroid  has 
lately  come  into  use  to  designate  it 


CENTRE    OF  (U  AV171\  .  .         V)V 

regarded  as  the  centre  of  parallel  forces,  it  is  more  truly 
conceived  of  as  the  "centre  of  mass;"  yet  in  deference  to 
usage  we  shall  call  the  point  the  "  centre  of  gravity." 

67.  Planes  of  Symmetry.— Axes  of  Symmetry. — If 

a  homogeneous  body  be  symmetrical  with  reference  to  any 
plane,  the  centre  of  gravity  is  in  that  plane. 

If  two  or  more  such  planes  of  symmetry  intersect  in 
one  line,  or  axis  of  symmetry,  the  centre  of  gravity  is  in 
that  axis. 

If  three  or  more  planes  of  symmetry  intersect  each  other 
in  a  point,  that  point  is  the  centre  of  gravity. 

By  observing  these  principles  of  the  symmetry  of  the 
figure  there  are  many  cases  where  the  centre  of  gravity  is 
known  at  once  ;  thus,  the  centre  of  gravity  of  a  straight 
line  is  its  middle  point.  The  centre  of  gravity  of  a  circle 
or  of  its  circumference,  or  of  a  sphere  or  of  its  surface,  is  its 
centre.  The  centre  of  gravity  of  a  parallelogram  or  of  its 
perimeter  is  the  point  in  which  the  diagonals  intersect. 
The  centre  of  gravity  of  a  cylinder  or  of  its  surface  is  the 
middle  of  its  axis ;  and  in  a  similar  manner  we  shall 
frequently  conclude  from  the  symmetry  of  the  figure,  that 
the  centre  of  gravity  of  a  body  is  in  a  particular  line  which 
can  be  at  once  determined. 

When  we  speak  of  the  centre  of  gravity  of  a  line,  we 
are  really  considering  a  material  line  of  the  same  density 
and  thickness  throughout,  whose  section  is  infinitesimal ; 
and  when  we  consider  the  centre  of  gravity  of  any  surface, 
wre  are  really  considering  the  surface  as  a  thin  uniform 
lamina,  the  thickness  of  which,  being  uniform,  can  be 
neglected. 

68.  Body  Suspended  from  a  Point—  When  a  body  is 
suspended  from  a  point  about  tvliich  it  can  turn  freely,  it 
will  rest  ivith  its  centre  of  gravity  in  the  vertical  line  passing 
through  the  point  of  suspension.     For,  if  the  point  of  sus- 


!02  BODY  SUPPORTED    ON  A   SURFACE. 

pension  and  the  centre  of  gravity  are  not  in  a  vertical  line, 
the  weight  acting  vertically  downwards  at  the  centre  of 
gravity  and  the  reaction  of  the  point  of  suspension  vertically 
upwards  form  a  statical  couple  and  hence  there  will  be 
rotation. 

69.  Body  Supported  on  a  Surface.—  When  a  body  is 
placed  on  a  surface  it  will  stand  or  fall  according  as  the 
vertical  line  through  the  centre  of  gravity  falls  within  or 
without  the  base.  For  if  it  falls  within  the  base  the  reaction 
of  the  surface  upward  and  the  action  of  the  weight  down- 
ward will  be  in  the  same  vertical  line,  and  so  there  will  be 
equilibrium.  But  if  it  falls  without  the  base  the  reaction 
of  the  surface  upward  and  the  action  of  the  weight  down- 
ward form  a  statical  couple  and  hence  the  body  will  rotate 
and  fall. 

70.  Different  Kinds  of  Equilibrium. — According  to 
the  proposition  just  proved  (Art.  69)  a  body  ought  to  rest 
upon  a  single  point  without  falling,  provided  that  its  centre 
of  gravity  is  placed  in  the  vertical  line  through  the  point 
which  forms  its  base.  And,  in  fact,  a  body  so  situated 
would  be,  mathematically  speaking,  in  a  position  of  equili- 
brium, though  practically  the  equilibrium  would  not  sub- 
sist. The  body  would  be  moved  from  its  position  by  the 
least  force,  and  if  left  to  itself  it  would  depart  further  from 
it,  and  never  return  to  that  position  again.  This  kind  of 
equilibrium,  and  that  which  is  practically  possible,  are 
distinguished  by  the  names  of  unstable  and  stable.  Thus 
an  egg  on  either  end  is  in  a  position  of  unstable  equilibrium, 
but  when  resting  on  its  side  it  is  in  a  position  of  stable 
equilibrium.  The  distinction  may  be  defined  generally  as 
follows : 

When  the  body  is  in  such  a  position  that  if  slightly  dis- 
placed it  tends  to  return  to  its  original  position,  the  equili- 
brium is  stable.    When  it  tends  to  move  further  away  from 


CENTRE     OF    GRAVITY    OF   A     TRIANGLE.  103 

its  original  position,  its  equilibrium  is  unstable.  When  it 
remains  in  its  new  position,  its  equilibrium  is  neutral.  A 
sphere  or  cylindrical  roller,  resting  on  a  horizontal  surface, 
is  in  neutral  equilibrium.  In  stable  equilibrium  the  centre 
of  gravity  occupies  the  lowest  possible  position;  and  in 
unstable  it  occupies  the  highest  position. 
We  shall  first  give  a  few  elementary  examples. 

71.  Given  the  Centres  of  Gravity  of  two  Masses, 
Ml  and  31 2,  to  find  the  Centre  of  Gravity  of  the  two 
Masses  as  one  System.— Let  gx,  denote  the  centre  of 
gravity  of  the  mass  Mlf  and  g2  the  centre  of  gravity  of  the 
mass  M2.    Join  gt  g2  and  divide  it  at  the  point,  G,  so  that 

-1  =  j~ ,  then  G  is  the  centre  of  gravity  of  the  two 

masses  as  one  system  (Art.  45). 

72.  Given  the  Centre  of  Gravity  of  a  Body  of 
Mass,  31,  and  also  the  Centre  of  Gravity  of  a  part 
of  the  Body  of  Mass,  m,  to  find  the  Centre  of 
Gravity  of  the  remainder. — Let  G  denote  the  centre  of 
gravity  of  the  mass,  M,  and  gx  the  centre  of  gravity  of  the 
mass,  mv  Join  Gg1  and  produce  it  through  G  tog2,  so  that 

jr1  =  y  _J — j  then  g2  is  the  centre  of  gravity  of  the 
remainder  (Art.  45). 

73.  Centre  of  Gravity  of  a  Triangular  Figure  of 
Uniform   Thickness  and   Density. — Let  ABC  be  the 

triangle;  bisect  BC  in  D,  and  join  AD; 

draw  any  line  bdc  parallel  to  BC  ;  then  it 

is  evident  that  this  line  will  be  bisected  by 

AD  in  d,  and  will  therefore  have  its  centre 

of  gravity  at  d;  similarly  every  line  in  the 

triangle  parallel  to  BC  will  have  its  centre 

of  gravity  in  AD,  and  therefore  the  centre  of  gravity  of  the 

triangle  must  be  somewhere  in  AD. 


104  CENTRE    OF    GRAVITY   OF   A     TRIANGLE. 

In  like  manner  the  centre  of  gravity  must  lie  on  the  line 
BE  which  joins  B  to  the  middle  point  of  AC.  It  is  there- 
fore at  the  intersection,  G,  of  AD  and  BE. 

Join  DE,  which  will  be  parallel  to  AB ;  then  the  triangles, 
ABG,  DE#,  are  similar;  therefore 

AG    _  AB  _  BO    _  2 

GD  ""  DE  ~~  DC  ~  1 ; 

or  GD  =  |AG  =  |AD. 

Hence,  to  find  the  centre  of  gravity  of  a  triangle,  "bisect  any 
side,  join  the  point  of  bisection  with  the  opposite  angle,  the 
centre  of  gravity  lies  one  third  the  way  up  this  bisection. 

Con.  1. — If  three  equal  particles  be  placed  at  the  vertices 
of  the  triangle  ABC  their  centre  of  gravity  will  coincide 
with  that  of  the  triangle. 

For,  the  centre  of  gravity  of  the  two  equal  particles  at  B 
and  C  is  the  middle  point  of  BO,  and  the  centre  of  gravity 
of  the  three  lies  on  the  line  joining  this  point  to  A. 
Similarly,  it  lies  on  the  line  joining  B  to  the  middle  of  AC. 
Therefore,  etc. 

Cor.  2. — The  centre  of  gravity  of  any  plane  polygon  may 
be  found  by  dividing  it  into  triangles,  finding  the  centre  of 
gravity  of  each  triangle,  and  then  by  Art,  59  deducing  the 
centre  of  gravity  of  the  whole  figure. 

Cor.  3. — Let  the  co-ordinates  of  A,  referred  to  any  axes, 
be  a?j,  y19  zt ;  those  of  B,  x2,  y2,  z2;  and  those  of  C,  xz, 
y3,  z3 ;  then  (Art.  59),  the  co-ordinates,  x,  y,  z,  of  the  centre 
of  gravity  of  three  equal  particles  placed  at  A,  B,  C,  respec- 
tively, are 

-  __  xt  +  a?8  -f  xs             Vi_±j2  +  V* 
•  - 3 ,    y- 3 , 


CENTRE   OF  GRAVITY   OF  A    PYRAMID.  105 

which  are  also  the  co-ordinates  of  the  centre  of  gravity  of 
the  triangle  ABC  (Cor.  1). 

74.  Centre  of  Gravity  of  a  Triangular  Pyramid  of 
Uniform  Density. — Let  D-ABC  be  a  triangular  pyramid; 
bisect  AC  at  E;  join  BE,  DE;  take  EF 
=  JEB,  then  F  is  the  centre  of  gravity  of 
ABC  (Art.  73).  Join  FD  ;  draw  ab,  be,  ca 
parallel  to  AB,  BC,  CA  respectively,  and 
let  DF  meet  the  plane,  abc,  at  /;  join  bf 
and  produce  it  to  meet  DE  at  e.     Then 

1  Fig. 38 

since  in  the  triangle  ADC,  ac  is  parallel 

to  AC,  and  DE  bisects  AC,  e  is  the  middle  point  of  ac;  also 

V-W  -3L. 

BF  ~  DF~~  EF' 
but  EF  =  |BF, 

therefore  ef=^bf; 

therefore  /is  the  centre  of  gravity  of  the  triangle  abc  (Art. 
73).  Now  if  we  suppose  the  pyramid  to  be  divided  by 
planes  parallel  to  ABC  into  an  indefinitely  great  number  of 
triangular  lamina?,  each  of  these  laminse  has  its  centre  of 
gravity  in  DF.  Hence  the  centre  of  gravity  of  the  pyramid 
is  in  DF. 

Again,  take  EH  =  JED ;  join  HB  cutting  DF  at  G. 
Then,  as  before  the  centre  of  the  pyramid  must  be  on  BH. 
It  is  therefore  at  the  intersection,  G-,  of  the  lines  DF 
and  BH. 

Join  FH  ;  then  FH  is  parallel  to  DB.  Also,  EF  =  |EB, 
therefore  FH  =  ^DB ;  and  in  the  similar  triangles,  FGH 
and  BGD,  we  have 

FG  __  F H  _  1  m 
DG  ~  DB  —  3 ' 

therefore  FG  =  JDG  =  £DF. 


106  CENTRE   OF  GRA  VITY  OF  A    CONE. 

Hence,  the  centre  of  gravity  of  the  pyramid  is  one-fourth 
of  the  loay  up  the  line  joining  the  centre  of  gravity  of  the 
base  ivith  the  vertex.  (Todhu liter's  Statics,  p.  108.  Also 
Pratt's  Mechanics,  p.  53.) 

Cor.  1. — The  centre  of  gravity  of  four  equal  particles 
placed  at  the  vertices  of  the  pyramid  coincides  with  the 
centre  of  gravity  of  the  pyramid. 

Cor.  2.— Let  (x  v  yt,  zx)  be  one  of  the  vertices;  (x2,y2,z2) 
a  second  vertex,  and  so  on;  let  (a,  y,  ~z)  be  the  centre  of 
gravity  of  the  pyramid  ;  then  (Art.  59) 

5  =  i  (xi  +  xz  +  ^3  +  X4>)> 

y  =  i  ivi  +  2/2  +  Vz  +  Vt)> 

*  =  i  (Zl    +  Z2    +   *3    +  Zj' 

Cor.  3. — The  perpendicular  distance  of  the  centre  of 
gravity  of  a  triangular  pyramid  from  the  base  is  equal  to  \ 
of  the  height  of  the  pyramid. 

75.  Centre  of  Gravity  of  a  Cone  of  Uniform 
Density  having  any  Plane  Base. — Consider  a  pyramid 
whose  base  is  a  polygon  of  any  number  of  sides.  Divide 
the  base  into  triangles;  join  the  vertex  of  the  pyramid  with 
the  vertices  of  all  the  triangles ;  then  we  may  consider  the 
pyramid  as  composed  of  a  number  of  triangular  pyramids. 
Now  the  centre  of  gravity  of  each  of  these  triangular 
pyramids  lies  in  a  plane  whose  distance  from  the  base  is 
one-fourth  of  the  height  of  the  pyramid  (Art.  74,  Cor.  3) ; 
therefore  the  centre  of  gravity  of  the  whole  pyramid  lies  in 
this  plane,  i.  e.9  its  perpendicular  distance  from  the  base  is 
one-fourth  of  the  height  of  the  pyramid. 

Again,  if  we  suppose  the  pyramid  to  be  divided  into  an 
indefinitely  great  number  of  laminae,  as  in  Art.  74,  each  of 
these  laminae  has  its  centre  of  gravity  on   the  right  line 


CENTRE    OF    GRAVITY.  107 

joining  the  vertex  to  the  centre  of  gravity  of  the  base;  and 
hence  the  centre  of  gravity  of  the  whole  pyramid  lies  on 
this  line,  and  hence  it  must  be  one-fourth  the  way  up  this 
line.  There  is  no  limit  to  the  number  of  sides  of  the  poly- 
gon which  forms  the  base  of  the  pyramid,  and  hence  they 
may  form  a  continuous  curve. 

Therefore,  the  centre  of  gravity  of  a  cone  whose  base  is 
any  plane  curve  whatever  is  found  by  joining  the  centre  of 
(/rarity  of  the  base  to  the  vertex,  and  taking  a  point  one- 
fourth  of  the  way  up  this  line. 

76.  Centre  of  Gravity  of  the  Frustum  of  a  Pyra- 
mid.— Let  ABC-abc  (Fig.  38)  be  the  frustum,  formed  by 
the  removal  of  the  pyramid,  D-abc,  from  the  whole  pyramid, 
D-ABC ;  let  hx  and  H  be  the  perpendicular  heights  of  these 
pyramids,  respectively;  let  m  and  M denote  their  masses; 
and  let  z19  z2,  z  denote  the  perpendicular  distances  of  the 
centres  of  gravity  of  the  pyramids  D-ABC,  and  D-abc,  and 
the  frustum,  from  the  base ;  then  we  have  (Art.  59,  Sch.  1) 


or 
But 


Also,  the  masses  of  the  pyramids  are  to  each  other  as  their 
volumes  *  by  (1)  of  Art.  10,  and  therefore  as  the  cubes  of 
their  heights.     Hence  (1)  becomes 

*  If  the  bodies  are  homogeneous,  the  volumes  or  the  weights  are  proportional  to 
tae  masses,  and  may  be  substituted  for  them. 


Mzt 

:z(M- 

■  m)  + 

mz2; 

8 

Mzt 

~~     M 

zt  = 

—  mz2 

—  m 

H 
4; 

108  EXAMPLES. 

h_jH*-{H-lhl)h1* 
•  Hs  —  V 

1    h*  -  ±Hhx*  +  3y 
-  H~]li    R2  +  2^i  +  3V 


(2) 


Instead  of  the  heights  we  may  use  any  two  corresponding 
lines  in  the  lower  and  upper  bases,  to  which  the  heights  are 
proportional,  as  for  example  AB  and  ab.  Denoting  these 
lines  by  a  and  b,  and  the  altitude  of  the  frustum  by  h,  (2) 
becomes 

_  h    a*  +  2ab  +  3ff» 
z  -  4  '    a2  +  ab  +  W  '  W 

This  is  true  of  a  frustum  of  a  pyramid  on  any  base,  a 
and  b  being  homologous  sides  of  the  two  ends,  and  hence  it 
is  true  of  the  frustum  of  a  cone  standing  on  any  plane  base. 

EXAMPLES. 

1.  Find  the  centre  of  gravity  of  a  trapezoid  in  terms  of 
the  lengths  of  the  two  parallel  sides,  a  and  b,  and  of  the 
line,  h,  joining  their  middle  points. 

Take  moments  with  reference  to  the  longer  parallel  side. 

Ans.  On  the  line  bisecting  the  parallel  sides  and  at  a 

h    a  +  2b 


distance  from  its  lower  end 


3     a  -f-  b 


2.  If  out  of  any  cone  a  similar  cone  is  cut  so  that  their 
axes  are  in  the  same  line  and  their  bases  in  the  same  plane, 
find  the  height  of  the  centre  of  gravity  of  the  remainder 
above  the  base. 

Take  moments  with  reference  to  the  base. 


INTEGRATION  FORMULAS.  109 

/ti  _  7/4 
Ans.  }•  y^ tt3,  where  A,  is  the  height  of  the  original 

cone,  and  h',  the  height  of  that  which  is  cut  out  of  it. 

3.  If  out  of  any  cone  another  cone  is  cut  having  the 
same  base  and  their  axes  in  the  same  line,  find  the  height 
of  the  centre  of  gravity  of  the  remainder  above  the  base. 

Ans,  %  (h  +  hi),  where  h  and  //x  are  the  respective 
heights  of  the  original  cone  and  the  one  that  is  cut  out 
of  it. 

4.  If  out  of  any  right  cylinder  a  cone  is  cut  of  the  same 
base  and  height,  find  the  centre  of  gravity  of  the  remainder. 

Ans.  |ths  of  the  height  above  the  base. 

77.    Investigations    Involving    Integration. — The 

general  formulae  for  the  co-ordinates  of  the  centre  of  gravity 
vary  according  as  we  consider  a  material  line,  an  area  or 
thin  lamina,  or  a  solid  ;  and  assume  different  forms  accord- 
ing to  the  manner  in  which  the  matter  is  supposed  to  be 
divided  into  infinitesimal  elements. 

In  either  case  the  principle  is  the  same ;  the  quantity  of 
matter  is  divided  into  an  infinite  number  of  infinitesimal 
elements,  the  mass  of  the  element  being  dm ;  multiplying 
the  element  by  its  co-ordinate,  x,  for  example,  we  get 
x  •  dm,  which  is  the  moment  of  the  element*  with  respect  to 
the  plane  yz  (Art.  63) ;  and  fx  •  dm  is  the  sum  of  the 
moments  of  all  the  elements  with  respect  to  the  plane  yz, 
and  which  corresponds  to  ZPx  of  Art.  63.  Also,  fdm  is 
the  sum  of  the  masses  of  all  the  elements  which  correspond 
to  ZP  of  the  same  Article.  Hence,  dividing  the  former  by 
the  latter  we  have 

*  The  moment  of  the  force  acting  on  element  dm  is  strictly  dm.g-x,  but  since 
the  constant  g  appears  in  both  terms  of  expression  f©r  co-ordinates  of  centre  of 
gravity,  it  may  be  omitted  and  it  becomes  more  convenient  to  speak  of  the  moment 
of  the  element,  meaning  by  it  the  product  of  the  mass  of  the  element  dm.  and.  its 
*rm,  x.  The  moment  of  an  element  measures  its  effect  in  determining  the  positiou 
of  the  centre  of  gravity. 


110  CENTRE   OF  GRAVITY  OF  A    LIXE. 

fx • dm  .  . 

x  =  -fM~  (1) 

Similarly  y  =  ^0,  (2) 

fz-dm.  .. 

the  limits  of  integration  being  determined  by  the  form  of 
the  body  ;  the  sign,  /,  is  used  as  a  general  symbol  of  sum- 
mation, to  be  replaced  by  the  symbols  of  single,  double,  or 
triple  integration,  according  as  dm  denotes  the  mass  of  an 
elementary  length  or  surface  or  solid.  Hence,  the  co-or- 
dinate of  the  centre  of  gravity  referred  to  any  plane  is  equal 
to  the  sum  of  the  moments  of  the  elements  of  the  mass 
referred  to  the  same  plane  divided  by  the  sum  of  the  elements, 
or  the  whole  mass.  If  the  body  has  a  plane  of  symmetry 
(Art.  67),  we  may  take  it  to  be  the  plane  xy,  and  only  (1) 
and  (2)  are  necessary.  If  it  has  an  axis  of  symmetry  we 
may  take  it  to  be  the  axis  of  x,  and  only  (1)  is  necessary. 

78.  Centre  of  Gravity  of  the  Arc  of  a  Curve.— If 

the  body  whose  centre  of  gravity  we  want  is  a  material  line 
in  the  form  of  the  arc  of  any  curve,  dm  denotes  the  mass  of 
an  elementary  length  of  the  curve. 

Let  ds  =  the  length  of  an  element  of  the  curve ;  let 
h  =  the  area  of  a  normal  section  of  the  curve  at  the  point 
(z,  y,  z),  and  let  p  =  the  density  of  the  matter  at  this 
point.  Then  (Art.  11),  we  have  dm  =  hpds,  which  is  the 
mass  of  the  element  ;  multiplying  this  mass  by  its  co-or- 
dinate, x,  for  example,  we  have  the  moment  of  the  element, 
(kpxdx),  with  respect  to  the  plane,  yz. 

Hence,  substituting  for  dm  in  (1),  (2),  (3),  of  Art.  77, 
the  linear  element,  l-pds,  we  obtain,  for  the  position  of  the 
centre  of  gravity  of  a  body  in  the  form  of  any  curve,  the 
equations 


EXAMPLES. 


Ill 


fhpxds 
fkpds  * 


(1) 


y  = 


fkpyds 
fkpds  '' 


(3) 


£  = 


fkpzds 
fkpds 


(3) 


The  quantities  k  and  p  must  be  given  as  functions  of  the 
position  of  the  point  (x,  y,  z)  before  the  integrations  can 
be  performed. 

If  the  curve  is  of  double  curvature  all  three  equations 
are  required.  If  it  is  a  plane  curve,  we  may  take  it  to  be 
in  the  plane  xy,  and  (1)  and  (2)  are  sufficient  to  determine 
the  centre  of  gravity,  since  s  =  0.  If  the  curve  has  an  axis 
of  symmetry,  the  axis  of  x  may  be  made  to  coincide  with 
it,  and  (1)  is  sufficient. 


EXAMPLES. 

1.  To  find  the  centre  of  gravity  of  a  circular  arc  of  uni- 
form thickness  and  density. 

Let  BC  be  the  arc,  A  its  middle  point, 
and  0  the  centre  of  the  circle.  Then  as 
the  arc  is  symmetrical  with  respect  to  OA 
its  centre  of  gravity  must  lie  on  this  line. 
Take  the  origin  at  0,  and  OA  as  axis  oi  x. 
Then,  since  k  and  p  are  constant,  (1)  be- 
comes 


x  = 


fxds 
Ids'' 


(1) 


x  being  the  co-ordinate  of  any  point,  P,  in  the  arc.  Let  0 
be  the  angle  POA,  and  a  the  radius  of  the  circle,,  and  let 
a  =  the  angle  BOA.     Then 


112  EXAMPLES. 

x  =  a  cos  0, 
and  ds  =  a  dd. 

Pa2  cos  0  d0  /""cos  6  dO 

„  -         t/_«  J  -a  sin  05 

Hence  a?  = — =  a — =  a « 


/a  rta 


de 


Therefore,  the  distance  of  the  centre  of  gravity  of  the  arc  of 
a  circle  from  the  centre  is  the  product  of  the  radius  and  the 
chord  of  the  arc  divided  by  the  length  of  the  arc. 

Cor. — The  distance  of  the  centre  of  gravity  of  a  semi- 

2a 
circle  from  the  centre  is  — • 
v 

2.  Find  the  centre  of  gravity  of  the  quadrant,  hi),  (Fig, 
39),  referred  to  the  co-ordinate  axes  OX,  OY. 
The  equation  of  the  circle  is 

x2  -f  y2  =  a2. 


dx  _     dy     _  vdx2  +  dy2  _ 


xds  = 


Vx2  +  y2 
ax  dx 


yds  =  a  dx, 

.,                                     ,         adx 
and  ds  = ; 

y 

which  in  (1)  and  (2),  after  canceling  h  and  p,  give 

Pa       X  dx  r  -,a 


2a 


ra      dx  r  .     ,xf 

/  —  sin-1  - 

«-'o   Vrt2  —  x2  L  alo 


,t 


EXAMPLES.  113 


„  ^2  _  ^2       L         tfJo 


3.  Find  the  centre  of  gravity  of  the  arc  of  a  cycloid. 

Take  the  origin  at  the  starting  point  of  the  cycloid,  and 
let  the  base  be  taken  as  the  axis  of  x.  The  equation  of 
the  curve  is 

x  =  a  vers-1  "  —  (2ay  —  y*)%  ; 


dx  _  dy  ds 

it  is  evident  that  the  centre  of  gravity  will  be  in  the  axis  of 
the  cycloid ;  therefore  x  =  na  ;  and  as  k  and  p  are  constant, 
(2)  becomes 

/*»    y&y 

_  Jo     (2a -#  __ 

y~      /*« Cly^ -!«• 

^o    (2a  —  y)* 
Cor. — For  the  arc  of  a  semi-cycloid,  we  get 
x  =  |«,     y  =  fa. 

4.  Find  the  centre  of  gravity  of  a  circular  arc  of  uniform 
section,  the  density  varying  as  the  length  of  the  arc  from 
one  extremity. 

Let  AB  (Fig.  39),  be  the  arc  ;  let  \i  be  the  density  at  the 
units  distance  from  A,  then  [is  will  be  the  density  at  the 
distance  s  from  A  ;  let  OA  be  the  axis  of  x,  and  a  the 
L_  AOB.  Then,  putting  [is  for  p,  and  a  cos  0,  a  sin  6,  a  dd, 
and  ad,  for  x,  y,  ds,  and  s,  in  (1)  and  (2), 


114  EXAMPLES. 

J  h  -  fJLCiO  •  a  cos  6  .  add  /   6  cos  d  dd 


si  2a 


I  k  '  iiad  •  add  I 

a  sin  «  -f  cos  a  —  1 


0^0 
0 


I  h*  pad    asm  d  .  a  dd  I  6  sin  6  dd 


V  = ?; =  a 


j  1c- pad  •  add  JO 


_    sin  a  —  a  cos  a 
=  2a s • 


Cor. — For  a  quadrant  we  get 

4a,        _x      -       8a 
•  =  ^(ir-2),     s,  =  _ 

5.  Find  the  centre  of  gravity  of  one-half  of  a  loop  of  a 
temniscate  whose  equation  is  r2  —  a2  cos  2d,  I  being  the 
length  of  the  half-loop. 

dr  rdd     _ds 

Uere  _a«  sin  20  ~  a2  cos  26  "  ^;  *%  etC* 

,         -         «2        -  ,**  —  1 

<4ftS.  «  =  -!-;      y  =  a2 ? — • 

2^Z  2*J 

6.  Find  the  centre  of  gravity  of  a  straight  rod,  the  den- 
sity of  which  varies  as  the  nth  power  of  the  distance  of 
each  point  from  one  end. 

Take  the  origin  at  this  end,  suppose  the  axis  of  x  to  coincide  with 
the  axis  of  the  rod,  and  let  I  =  the  length  of  the  rod. 

Ans.  <c  =  — ~  I. 
n  -f-  2 


CENTRE    OF  GRAVITY  OF  AX  AREA.  115 

7.  Find  the  centre  of  gravity  of  the  arc  of  a  semi-car* 
dioid,  its  equation  being 

r  =  a  (1  -f-  cos  6). 

Ans.  The  co-ordinates  of  the  centre  of  gravity  referred 
to  the  axis  of  the  curve  and  a  perpendicular  through  the 
cusp,  as  axes  of  x  and  y,  are 

as  =  y  =  $0. 

79.  Centre  of  Gravity  of  a  Plane 
Area. — Let  ABCD  be  an  area  bounded 
by  the  ordinates,  AC  and  BD,  the  curve 
AB  whose  equation  is  given,  and  the  axis 
of  x  ;  it  is  required  to  find  the  centre  of 
gravity  of  this  area,  the  lamina  (Art.  67) 
being  supposed  of  uniform  thickness  and  density.  We 
divide  the  area  into  an  infinite  number  of  infinitesimal 
elements  (Art.  77).  Suppose  this  to  be  done  by  drawing 
ordinates  to  the  curve.  Let  PM  and  QN  be  two  consecu- 
tive ordinates,  let  (x,  y)  be  the  point,  P,  and  let  g  be  the 
centre  of  gravity  of  the  trapezoid,  MPQN,  whose  breadth  is 
dx  and  whose  parallel  sides  are  y  and  y  +  dy.  The  area  of 
this  trapezoid  is  y  dx,  (Gal.,  Art.  184). 

Let  p  be  the  density  and  h  the  thickness  of  the  lamina. 
Then  (Art.  11)  we  have  dm  =  kpy  dx,  which  is  the  mass 
of  the  element  MPQN;  multiplying  this  mass  by  its  co-or- 
dinate, x,  for  example,  we  have  the  moment  of  the  element 
(kfj  xy  dx),  with  respect  to  OY,  and  multiplying  by  the 
other  co-ordinate,  \y,  we  have  the  moment  with  respect  to 
OX.  Hence,  substituting  for  dm  in  (1)  and  (2)  of  Art,  77, 
the  surface  element,  kpy  dx,  and  remembering  that  k  and  p 
are  constants,  we  obtain,  for  the  position  of  the  centre  of 
gravity  of  a  body  in  the  form  of  a  plane  area,  the  equations, 


116  EXAMPLES. 

X~   fydx>     y-$J'ydx>  (1» 

the  integrations  extending  over  the  whole  area  CABD. 

EXAM  PLES. 

1.  Find  the  centre  of  gravity  of  the  area  of  a  semi-parab 
ola  whose  equation  is  y2  =  2px. 

Let  a  =  the  axis,  and  b  the  extreme  ordinate,  then  we 
have  from  (1) 

/    V%px?dx         I   x%  dx 

W  =     Pa     —    x        ==      f*a   l         ~~  *°  * 
I    v%p  x'z  dx         /    xs  dx 


pa  pa 

2pxdx  r    I   xdx 

«  -  1    Jo  -  A  /P  J°  -  ih 

9  ""   *      (*     r-     X  ""  V     2      /*    JL  ""  * 

/    V2px^dx  v      J   x^dx 

2.  Find  the  centre  of  gravity  of  the  area  of  an  elliptic 
quadrant  whose  equation  is 


=  -  V«2  —  &. 
a 


Here 


I  xy  dx         I    -  (a2  —  x2)^  x  dx 
faydx         fa-(a*-x2)*dx 


4a 
3^5 


ry2dx  rl-Aa2-x*)dz 

*  -  r°        -  ^°  a 

1  —  ?     pa  ~   2     pah  t       * 

£yd*     f-ai* -*)**> 


EXAMPLES.  117 

-  U 

•'•    *  =  Tn 

Hence  for  the  centre  of  gravity  of  the  area  of  a  circulai 
quadrant  we  have 

_       4a 

•-'-fir 

3.  Find  the  centre  of  gravity  of  the  area  of  a  sem; 
cycloid. 

Take  the  axis  of  the  curve  as  axis  of  x,  and  a  tangent  at 
the  highest  point  as  axis  of  y ;  then  the  equation  is  (Anal. 
Geom.,  Art.  157), 


x 


y  =  a  vers-1  -  -f-  V^ax  —  x2; 
ci 

where  a  is  the  radius  of  the  generating  circle.    From  (1)  we 
have 

__70  **__]?% -J s*i 

x  /»2a  r-  n  -i2a 

[yx  -  f{%ax  -  *)*  dx  J       '?"■*-  £™8 

since  when  #  =  0  and  2a,  y  =  0  and  m 

•  *•     a?  =  $#. 

Also, 

J    y2 dx       [tfx  —  %  J  yx dy\^ 

</0   '** 


118 


POLAR   ELEMENTS    OF  A   PLANE  AREA, 


y2x—2   I  y  (2ax  —  x1)^  dx  V 


Sna2 


y2x—2a  I  (2ax—x2)i  vers-1  -  dx— 2  I  (2ax—x%)  dx  | 


3™2 


y2x  —  2ax2  4-  -fa-3  —  2a  I  (2ax  —  x2)%  vers-1  -  dx] 


3na? 

2tt*o? 

8a3 
3 

A3 

~2~ 

f  A3  - 
3rra2 

8a3 
3 

3na? 

9 

•  • 

y  = 

-^- 

•I). 

which  the  student  can  verify  by  assuming 


vers-1  -  =  0. 
a 


(See  Todhunter's  Statics,  p.  118.) 

80.  Polar  Elements  of  a  Plane 
Area. — Let  AB  be  the  arc  of  a  curve, 
and  let  it  be  required  to  find  the  centre  of 
gravity  of  the  area  bounded  by  the  arc 
AB  and  the  extreme  radii-vectors,  0A 
and  OB,  drawn  from  the  pole,  0,  to  the 
extremities  of  the  arc. 

Divide  the  area  into  infinitesimal  triangles,  such  as  POQ, 
included  between  two  consecutive  radii-vectors,  OP  and 
OQ.  Let  (r,  6)  be  the  point,  P,  then  the  area  of  the 
element,  POQ  =  ±r2  dd  (Cal.,  Art.  191)  ;  and  if  the  thick- 
ness and  density  of  the  lamina  are  uniform,  the  centre  of 


Fig.41 


EXAMPLE.  119 

gravity  of  this  elementary  triangle  will  be  on  a  straight  line 
drawn  from  0  to  the  middle  of  PQ,  and  at  a  distance  of 
two-thirds  of  this  straight  line  from  0  (Art.  73).  Hence 
the  co-ordinates  of  the  centre  of  gravity,  g,  of  POQ,  are 
OM  and  Ug,  or, 

f  r  cos  0,     and    §r  sin  0. 

Hence,  (Art  77), 

/jr  cos  0  .  ^r2  dO  _      fr*  cos  0d0m  . 

*~~  fir*  dO  —  *      fr*dS      5  I1' 

fir  sin  0  *  |r2  ^0  _     >V3  sin  0  jg  #  . 

y"  fir*  dO  "~*     /r2tf0      '  W 

the  integrations  extending  over  the  whole  area,  AOB. 

e  x  AM  PLE, 

Find  the  centre  of  gravity  of  the  area  of  a  loop  of  Ber- 
noulli's Lemniscate  whose  equation  is  r2  ==  a2  cos  20. 

As  the  axis  of  the  loop  is  symmetrical  with  respect  to 
the  axis  of  x,  y  =  0,  and  the  abscissa  of  the  centre  of 
gravity  of  the  whole  loop  is  evidently  the  same  as  that  of 
the  half-loop  above  the  axis.  Substituting  in  (1)  fir  r  its 
value  a  cos^  20,  we  have 


coss  20  COS  0 


/%os  20  dd 
s  taj*(l  —  2  sin3  0)t  d  sinfc 


Put  sin  0  =  — ^? ,  theu 

V2 


120  DOUBLE  INTEGRATION. 


s  =  .i^L   /%0S4  cf>d<i>  =  —?-  •  1 1  (Cal.,  Art.  157) 

81.  Double  Integration.— Polar  Formulae.—  When 
the  density  of  the  lamina  varies  from  point  to  point,  it  may 
be  necessary  to  divide  it  into  elements  of  the  second  order 
instead  of  rectangular  or  triangular  elements  of  the  first 
order  (Arts.  79  and  80). 

Suppose  that  the  density  of  the  lamina  AOB  (Fig.  41), 
is  not  uniform.  If  we  divide  it  into  triangular  elements, 
POQ,  the  element  of  mass  will  be  no  longer  proportional  to 
the  element  of  area,  POQ  =  fr2  dd ;  nor  will  the  centre  of 
gravity  of  the  triaugle,  POQ,  be  \r  distant  from  0. 

Let  a  series  of  circles  be  described  with  0  as  a  centre, 
the  distance  between  any  two  successive  circles  being  dr. 
These  circles  will  divide  the  triangle,  POQ,  into  an  infinite 
number  of  rectangular  elements,  abed  =  rdd  dr.  If  lc  is 
the  thickness  and  p  is  the  density  of  the  lamina  at  this  ele- 
ment, the  element  of  mass  will  be  d?n  =  k  pr  dd  dr  ;  and 
the  co-ordinates  of  its  centre  of  gravity  will  be  r  cos  0  and 
r  sin  6.     Hence,  from  (1)  and  (2)  of  Art.  77,  we  have 


x  = 


/   /  k  pr  cos  0  rdd  dr         II  kpr*  cos  6  dd  dr 
Tfkpr  dd  dr  J  J  kpr  dd  dr 


(1) 


f  fkpr2  sin  d  dd  dr 

and  y  =  J-^? (2) 

/   /  kpr  dd  dr 

In  each  of  these  integrals  the  values  of  k  and  p  are  to  be 
substituted  in  terms  of  r  and  0}  and  the  integrations  taken 
between  proper  limits. 


EXAMPLE.  121 


EXAMPLE. 


Find  the  centre  of  gravity  of  the  area  of  a  cardioid  in 
which  the  density  at  a  point  increases  directly  as  its  distance 
from  the  cusp. 

Let  \i  =  the  density  at  the  unit's  distance  from  the 
cusp,  then  p  ==  \ir,  is  the  density  at  the  distance  r  from 
the  cusp. 

As  the  axis  of  the  curve  is  an  axis  of  symmetry  (Art.  67), 
y  =  0,  and  the  abscissa  of  the  whole  curve  is  the  same  as 
for  the  half  above  the  axis ;  then  (1)  becomes 


rs  cos  9  dd  dr 
x  = 


=  i 


r  frr2  dd  dr 
I   r*  cos  0  dd 


I 

by  performing  the  r-integration. 
The  equation  of  the  curve  is 


r  =  a  (1  +  cos  0)  =  2a  cos2  jr« 


Substituting  this  value  for  r,  and  putting  -  =  (f>,  we  have 


x  =  f  a 


/%os8  <b  (2  cos2  $  —  1)  d<f> 


I    cos6  (p  d<f> 


122  RECTANGULAR    FORMULAE. 

82.  Double  Integration. — Rectangular  Formulae.— 

Let  a  series  of  consecutive  straight  lines  be  drawn  parallel 
to  the  axes  of  x  and  y  respectively,  dividing  the  area,  ABCD, 
(Fig.  40),  into  an  infinite  number  of  rectangular  elements 
of  the  second  order.  Then  the  area  of  each  element,  as 
abed,  =  dxdy;  and  if  k  and  p  are  the  thickness  and  density 
of  the  lamina  at  this  element,  the  element  of  mass  will  be 
dm  =  kp  dx  dy ;  and  the  co-ordinates  of  its  centre  of  gravity 
will  be  x  and  y.  Hence  from  (1)  and  (2)  of  Art.  77,  we 
have 

/   /Is  px  dx  dy 

%  =  ±-± ;  (1) 

/ jkp  d®  dy 

I  I  hpy  dx  dy 

V  =  ^t~ (8) 

/    /  kp  dx  dy 

the  integrations  being  taken  between  proper  limits. 

EX A  M  PLE 

Find  the  centre  of  gravity  of  the  area  of  a  cycloid  the 
density  of  which  varies  as  the  nth  power  of  the  distance 
from  the  base. 

Take  the  base  as  the  axis  of  x  and  the  starting  point  aa 
the  origin.     Then  the  equation  of  the  curve  is 


x  =  a  vers'1  -  —  (2ay  —  ^)^ ; 


.  • .    dx  = 


_      yty 


V%ay  —  y2 


SURFACE    OF    REVOLUTION.  123 

Let  p  =  fiyn  =  density  at  the  distance  y  from  the  base. 
It  is  evident  that  the  centre  of  gravity  will  be  in  the  axis  of 
the  cycloid  ;  therefore  x  =  na  ;  and  as  k  is  constant  (2) 
becomes 

J0     J^dydx 

y  = 


/ 

n  -f-  l^o 


J      yn+1dx 
yn+zdy 


V%«y  —  y2 


n  +  2     /*a       yn^dy 

Jo      */~2av  — 


V%ay  —  y 


J 

+  5     ^o 


!/' 


_  n  +  1    2n  +  5     ^o     V2«y  —  y2  m 
~"  ft  +  2  '  w  +  3  fl  .  r**     yn+*dy     ' 
*^o      V%ay~—f 

.  _  w  -f  1    2n  +  5 

83.  Centre  of  Gravity  of  a  Surface  of  Revolu- 
tion.--Let  a  surface  be  generated  by  the  revolution  of  the 
curve,  AB  (Fig.  40),  round  the  axis  of  x.  Then  the 
elementary  arc,  PQ,  (=  ds),  generates  an  element  of  the 
surface  whose  area  =  *2ny  ds  (CaL,  Art.  193).  If  k  is  the 
thickness  and  p  the  density  of  the  lamina  or  shell  in  this 
elementary  zone,  the  element  of  mass  will  be  dm  =  2nkpy  ds. 
Also  the  centre  of  gravity  of  this  zone  is  in  the  axis  of  x  at 


124  EXAMPLES. 

the  point  if  whose  abscissa  is  x  and  ordinate  0.    Henc*  (1) 
of  Art.  77  becomes,  after  cancelling  2n, 


I  kpxy  ds 

J—r —  (i) 


the  integrations  being  taken  between  proper  limits. 


examples. 

1.  Find  the  centre  of  gravity  of  the  surface  formed  by 
the  revolution  of  a  semi-cycloid  round  its  base. 
The  equation  of  the  generating  curve  is 

x  =  a  vers-1  ^  —  V^ay  —%?} 


dx  _  dy  _      ds 

V  "  V2ay  —  f  ~~  V2ay' 


or  ds  = 


V2ady 


V2a  —  y 

which  in  (1)  gives,  after  cancelling  V^a  hp9 
/**    xydy 

„—  °  ^a  —  y 

ydy 


f 


«.      .7AT-H* 


V2a  —  y 


2.  Find  the  centre  of  gravity  of  the  surface  formed  by 
the  revolution  of  a  semi-cycloid  round  its  axis. 

It  is  clear  that  the  centre  of  gravity  lies  on  the  axis  of 
the  curve ;  hence  y  =  0. 


EXAMPLES,  125 

The  equation  of  the  generating  curve  is 


v  =:  a  vers-1  -  -4-  V  2ax  —  a& 

•  a 


Here  -       dy  =  V^^<*°> 

X 

ds  =  V2ax~»  dz, 


which  in  (1)  gives 


/    yx%  dx 

x  = 


\\yx%  —  \  I  xV%a  —  x  dx~T 
hyx?  —  2fV2a  —  x  dxf 

"  2na  {2a)?  -  |  (2a)* 
„        15rr  —  8 

3.  Find  the  centre  of  gravity  of  the  surface  formed  by 
the  revolution  of  the  semi-cycloid  round  the  axis  of  y  in  the 
last  example,  t.  e.,  round  the  tangent  to  the  curve  at  the 
highest  point. 

Ans.  y  =  --  (15tt  —  8). 


126  ANT    CURVED    SURFACE. 

84.  Centre  of  Gravity  of  Any  Curved  Surface.— 

Let  there  be  a  shell  having  any  given  curved  surface  for 
one  of  its  boundaries;  and  let  k  =  the  thickness,  p  =  the 
density,  and  ds  =  the  area  of  an  element  of  the  surface  at 
the  point  (x,  y,  z) ;  then  (1)  of  Art.  83  becomes 


/  kpx  ds 

J—r —  (i) 

/  kp  ds 


and  similar  expressions  for  y  and  z. 

Substituting  the  value  of  ds  (Cal.,  Art.  201)  and  cancel, 
ling  k  and  p,  we  have 


m 


JJ[1  +  a*  +  df)  dxdy 


EXAMPLES. 


1.    Find    the  centre  of    gravity  of  one-eighth  of    the 
surface  of  a  sphere. 

Here  x*  -f  f  -f  z2  =  <x\ 

\    +  dx2  +  df)    ~  («a  —  x2  —  #3)*' 
/■  /*       xdxdy 


SOLID    OF  REVOLUTION. 


w 


First  perform  the  y-integra- 
tion,  x  being  constant,  from 
y  ==  0  to  y  =  LI  =  y1  = 
Va2  —  x* ;  the  effect  will  be 
to  sum  up  all  the  elements 
similar  to  pq  from  H  to  I. 
The  effect  of  a  subsequent 
^--integration  will  be  to  sum 
all  these  elemental  strips  that 
are  comprised  in  the  surface 
of  which  OAB  is  the  projec- 
tion, and  the  limits  of  this  integration  are  x  =  0  and 
x  =  OA  =  a.    Hence 


Fig.  42. 


x  = 


yvvi 


#  rfo  % 


.V'2 


V 


5* 


Jo    e/0      (a2  _  38  _  ^i 

/ 


-§-7r#  dx 


ffindx 


=    J«. 


Similarly 


y  =  }ff,     s  =  Jtf. 


2.   Find  the  centre  of  gravity  of  one-eighth  of  the  surface 

of  the  sphere  if  the  density  varies  as  the  ^-ordinate  to  any 

point  of  it.     Here  p  =  \iz. 

.  4tf     -        4a  2a 

Ms.  »  =  ^;  y  =  -;  8  =  y. 


85.  Centre  of  Gravity  of  a  Solid  of  Revolution. — 

Let  a  solid  be  generated  by  the  revolution  of  the  curve,  AB, 
(Fig.  40),  round  the  axis  of  x.  Then  the  elementary 
rectangle,  PQNM,  (—  ydx),  generates  an  element  of  the 


128  SOLID    OF  I? EVOLUTION. 

solid  whose  volume  =  ny2  dx  (Cal.,  Art.  203).  Hence  if  the 
density  of  the  solid  is  uniform,  we  have  for  the  position  of 
the  centre  of  gravity  (which  evidently  is  in  the  axis  of  x), 


j  ny2x  dx         I  y2x  dx 
I  ny2  dx  I  y2  dx 


a) 


the    integrations    being    extended    over    the  whole  area, 
CABD,  of  the  bounding  curve. 

If  the  density  varies,  the  element  of  mass  may  require  to 
be  taken  differently.  If  the  density  varies  with  x  alone,  i.  e., 
if  it  is  uniform  all  over  the  rectangular  strip,  PQNM,  the 
volume  may  be  divided  up  as  already  done,  and  the  element 
Df  mass  =  irpy2  dx.     Hence,  we  shall  have  in  this  case, 

/  py2x  dx 

i  =  v—  w 

J  py2  dx 

If  the  density  varies  as  y  alone,  we  may  take  a  rectangular 
element  of  area  of  the  second  order,  dx  dy,  at  the  point 
(x,  y)  ;  this  area  will  generate  an  element  of  volume 
=  2ny  dx  dy  ;  therefore  the  element  of  mass  =  %npy  dx  dy, 
and  we  have 

f  Cpxydxdy 

(3) 


J J pydxdy 


the  ^-integrations  being  performed  first,  from  0  to  y,  the 
ordinate  of  a  point  P,  on  the  bounding  curve ;  and  then 
the  ^-integrations  from  OC  to  OD. 


EXAMPLES.  129 

EXAMPLES. 

1.  Find  the  centre  of  gravity  of  the  hemisphere  generated 
by  the  revolution  of  the  quadrant,  AD,  (Fig.  39),  round  OA 
(taken  as  axis  of  x),  (1)  when  the  density  is  uniform ;  (2) 
when  it  is  constant  over  a  section  perpendicular  to  OA  and 
varies  as  the  distance  of  this  section  from  OD ;  (3)  when 
it  is  constant  at  the  same  distance  from  OA  and  varies  as 
this  distance. 

(1)  From  (1)  we  have 

/  y2x  dx 

x=  —ji 

/  y2dx 

Putting  x  =  r  cos  0,   and   y  =  r  sin  0,   where  r  is   the 
radius  of  the  circle    and  integrating  between  0  =  0  and 

0  =  - ,  we  have 

*  =  fr. 

(2)  Since  p  =  fix,  we  have  from  (2) 

/  x2y2  dx 

•  =  -7 > 

J  xy2  dx 

which  gives  x  =  -£$r, 

(3)  Since  p  =  \iy,  we  have  from  (3) 

/    /  xy2  dx  dy         I  xyz  dx 


«  = 


I       y2  dx  dy  I  y*  d-i 


130  EXAMPLES. 

and  the  previous  substitutions  for  x  and  y  give 
.  _  16r 

X  ~~    157T* 

2.  Find  the  centre  of  gravity  of  a  paraboloid  of  revolu- 
tion, the  length  of  whose  axis  is  h.  Ans.  x  =r  \h. 

3.  Find  the  centre  of  gravity  (1)  of  a  portion  of  a  prolate 
spheroid,  the  length  of  whose  axis  measured  from  the  vertex 
is  c,  and  (2)  of  a  hemi-spheroid. 

.        ,nX  -       c  8a  —  3c    /nl  _        _ 
^s.  (1)  s  =  -  -3^7--;  (2)  *  =  fa. 

86.  Polar  Formulas. — Let  a  solid  be  generated  by  the 
revolution  of  AB,  (Fig.  41),  round  the  axis  of  #.  Then  the 
elementary  rectangle,  abed,  whose  mass  =  pr  dd  dr,  (Art. 
81),  the  thickness  being  omitted,  generates  a  ring  which  is 
an  element  of  the  solid  whose  volume  =  2nr  sin  6  pr  dd  dr; 
and  the  abscissa  of  the  centre  of  gravity  of  the  ring  is 
r  cos  6,    Hence  (1)  of  Art.  77  becomes 

/   /  pr5  sii 

f  fpr2  sin  6  dd  dr 

in  which  p  must  be  a  function  of  r  and  6  in  order  that  the 
integrations  may  be  effected. 

If  the  density  depends  only  on  the  distance  from  a  fixed 
point  in  the  axis  of  revolution,  this  point  may  be  taken  as 
origin,  and  p  will  be  a  function  of  r  ;  if  the  density  depends 
only  on  the  distance  from  the  axis  of  revolution,  p  will 
be  a  function  of  r  sin  6. 

EXAMPLE. 

The  vertex  o»  a  right  circular  cone  is  in  the  surface  of  a 
sphere,  the  axis  of  the  cone  coinciding  with  a  diameter  of 


sin  6  cos  0  dd  dr 

(i) 


CENTRE   OF  GRAVITY  OF  ANY  SO  LTD.  I'M 

the  sphere,  the  base  of  the  cone  being  a  portion  of  the  sur- 
face of  the  sphere.  Find  the  distance  of  the  centre  of 
gravity  of  the  cone  from  its  vertex,  2«  being  its  vertical 
angle,  and  a,  the  radius  of  the  sphere. 

Here  the  r-limits  are  0  and  2a  cos  0 ;  the  0-limits  are  0 
and  a ;  p  is  constant ;  hence  from  (1)  we  have 


«  = 


pa    pr 

I     /    r3  sin  0  cos  0  dd  dr 

«'0    'r0 


•2  sin  0  dd  dr 


^i 


=  |« 


/    (2a  cos  0)4  sin  0  cos  0  d$ 
«{o m 

fU(2a  cos  0) 
/™cos5  0  sin  0  rffl 


sin  0  dd 


/""cos3  0  sin  0  d9 
—  cos6  « 


1  —  cos4  a 

87.  Centre  of  Gravity  of  any  Solid.— Let  (x,  y,  z) 

and  (x  -+-  dx,  y  -f  dy,  z  -f-  dz)  be  two  consecutive  points  E 
and  F,  (Fig.  42),  within  the  solid  whose  centre  of  gravity  is 
to  be  found.  Through  E,  pass  three  planes  parallel  to  the 
co-ordinate  planes  xy,  yz,  zx;  also  through  F  pass  three 
planes  parallel  to  the  first.  The  solid  included  by  these  six 
planes  is  an  infinitesimal  parallelopiped,  of  which  E  and  F 
are  two  opposite  angles,  and  the  volume  =  dx  dy  dz.  If  p 
is  the  density  of  the  body  at  E,  the  element  of  mass  at  E 
=3  o  dx  dy  dz.  Hence  the  co-ordinates  of  the  centre  of 
gravity  of  the  solid  are  given  by  the  equations 


(1) 


(2) 


(3) 


132  EXAMPLES, 

I    I    I  px  dx  dy  dz 

3  =  ~FTT — " * 

jjl?  dx  dy  dz 

tfjtfpydxdydz 
J   I   /  pdxdydz 

/    I    I  pz  dx  dy  dz 
9  dx  dy  dz 

the  integrations  being  extended  over  the  whole  solid. 

EXAMPLES. 

1.  Find  the  centre  of  gravity  of  the  eighth  part  of  an 
ellipsoid  included  between  its  three  principal  planes.* 

Let  the  equation  of  the  ellipsoid  be 

x2      y2      z2 1 

«2  +  p  T  •?  -  L 

Here  the  limits  of  the  ^-integration  are 

•(i-S-fl*-* 

which  call  zt  and  0  ;  the  limits  of  y  are 

LI  =  I  (l  -  D*  and  °> 
which  call  yx  and  0 ;  the  z-limits  are  a  and  0. 


*  Planes  of  xy,  yz,  zx. 


POL  Ah'   ELEMENTS   OF  MASS.  133 


First   integrate    with   respect  to   z,  and    we  obtain    the 

infinitesimal  prismatic  column  whose  base  is  PQ,  (Fig.  V2 ;, 
and  whose  height  is  Pp.  Then  we  integrate  with  respect 
to  y,  and  obtain  the  sum  of  all  the  columns  which  form 
the  elemental  slice  Kphnq.  Then  integrating  with  respect 
to  x,  we  obtain  the  sum  of  all  the  slices  included  in  the 
solid,  OABC.  Hence  (1)  becomes,  since  the  density  is 
uniform, 

/     I      jx  dx  dy  dz 
*^o  t7o    ^o 

/     /      /   \lxdydz 
VQ  t/o    ^o 

■OH-S-ff** 


a?  = 


r('-3 


6?a? 


«  =  f«. 


Similarly  y  =  \b,     z  =  §c. 

2.  Find  tht  centre  of  gravity  of  the  solid  bounded  by  the 
planes  z  =  $x,  z  ^  yx,  and  the  cylinder  y2  =  'lax  —  a;2. 

Jji*.  x  =  fa;  y  =  0;  £  =  —  (/3  +  y). 

88.  Polar  Elements  of  Mass.— Let  Fig.  43  repre- 
sent the  portion  of  the  volume  of  a  solid  included  between 
its  bounding  surface  and  three  rectangular  co-ordinate 
planes. 


134 


POLAR   ELEMENTS   OF  MASS. 


(1)  Through  the  axis  of  z  draw 
a  series  of  consecutive  planes,  divid- 
ing the  solid  into  wedge-shaped 
slices  such  as  COBA. 

(2)  Round  the  axis  of  z  describe 
a  series  of  right  cones  with  their 
vertices  at  0,  thus  dividing  each 
slice  into  elementary  pyramids  like 
O-PQST. 

(3)  With  0  as  a  centre  describe 
a  series   of    consecutive  spheres; 
thus  the  solid  is  divided  into  elementan  rectangular  par- 
allelopipeds  similar  to  abpt,  whose  volume  =  ap  -ps  •  sL 

Let  XOA  =  $,     COP  =  6,     Op  —  r, 

AOB  =  d<t>,  POQ  =  dO,  pa  =  dr. 

Then  pq  is  the  arc  of  a  circle  whose  radius  is  r,  ?pd  the 
angle  is  dd  ;  therefore 

pq  =  rdB. 

Also  ps  is  the  arc  of  a  circle  in  which  the  angle  is  d(p, 
and  the  radius  is  the  perpendicular  from  p  on  OZ,  or 
r  sin  6 ;  therefore 

ps  =  r  sin  6  d<f>. 

Therefore  the  volume,  of  the  elementary  parallelopiped  = 

r2  sin  6  dr  ddd<f>; 

and  if  p  is  the  density  of  the  solid  at  p,  the  element  of 
mass  is 

pr2  sin  6  dr  dd  dp. 

Also   the  co-ordinates  of  the   centre   of  gravity  of    this 
element  are 


r  sin  6  cos  $,     r  sin  6  sin  0,     and    r  cos  6 ; 


EXAMPLES.  135 

hence  for  the  centre  of  gravity  of  the  whole  solid  we  have 
/    /  J  p?'3  sin2  0  cos  (p  dr  dd  d(f> 


x  = 


/  f  'J'pr2  sin  6  dr  dO  d(f> 

f  C  Ppr*  sin2  0  sin  0  dr  dd  d<J> 
fjfpr2  sin  e  dr  dd  d<j> 

III  Pr*  s^n  0  cos  0  dr  dO  d<p 


P  = 


9  = 


fffpr2  sili  d  dr  dd  d<f> 


the  limits  of  integration  being  determined  by  the  figure  of 
the  solid  considered. 

The  angles,  6  and  0,  are  sometimes  called  the  co-latitude, 
and  longitude,  respectively. 


EXAM  PLES, 

1.  Find  the  centre  of  gravity  of  a  hemisphere  whose 
density  varies  as  the  wth  power  of  the  distance  from  the 
centre. 

Take  the  axis  of  z  perpendicular  .to  the  plane  base  of  the 
hemisphere.  Let  a  =  the  radius  of  the  sphere,  and 
p  =  fir*,  where  ^  is  the  density  at  the  units  distance  from 
the  centre.  First  integrate  with  respect  to  r  from  0  to  a, 
and  we  obtain  the  infinitesimal  pyramid  O-PQST.  Then 
integrate  with  respect  to  0  from  0  to  Jt,  and  we  obtain  the 
sum  of  all  the  pyramids  which  form  the  elemental  slice, 
COB  A.  Then  integrating  with  respect  to  (p  from  0  to  2n, 
we  obtain  the  sum  of  all  the  slices  included  in  the  hemi- 
sphere.    Hence, 


136  SPECIAL   METHODS. 


I       /    rn+3  sin  9  cos  6  dr  eld  dfi 

'-'p       "Q 

/      /      /    r»+2sin  6drd6d<j> 

*SQ        VQ        t/o 


w  +  3 

7—ia 

n  +  4 


/       /     sin  0  cos  0  f70  <70 
^o    «/o . 

/       /     sin  0  dd  d(f> 


0     ^0 

n  +  3    « 

and  it  is  clear  that        «  =  y  =  0. 

2.  Find  the  centre  of  gravity  of  a  portion  of  a  solid 
sphere  contained  in  a  right  cone  whose  vertex  is  the  centre 
of  the  sphere,  the  density  of  the  solid  varying  as  the  nth. 
power  of  the  distance  from  the  centre,  the  vertical  angle  of 
the  cone  being  =  2«,  and  the  radius  =  a. 

Take  the  axis  of  the  cone  as  that  of  z,  and  any  plane  through  it  as 
that  from  which  longitude  is  measured. 

Ans.  z  = r  «  (1  +  cos  a),  and  %  =  y  =  0. 

n  ■+-  4  2  V  y 

89.  Special  Methods. — In  the  preceding  Articles  we 
have  given  the  usual  formulae  for  finding  the  centres  of 
gravity  of  bodies,  but  particular  cases  may  occur  which  may 
be  most  conveniently  treated  by  special  methods. 

EXAMPLES. 

1.  A  circle  revolves  round  a  tangent  line  through  an 
angle  of  180°  ;  find  the  centre  of  gravity  of  the  solid 
generated. 


EXAMPLES. 


13? 


Let  OY  bo  the  tangent  line  about 
which  the  circle  revolves,  and  let  the 
plane  of  the  paper  bisect  the  solid ;  the 
centre  of  gravity  will  therefore  lie  in 
the  axis  of  x.  Let  P  and  Q  be  two 
consecutive  points  ;    and  let  OM  =  x, 

and    MP  =  y  =  V%ax  —  x\      The 
elementary  rectangle,  PQqp,  will  gen- 
erate a  semi-cylindrical   shell,  whose  volume  =  2ynxdx, 
the  centre  of  gravity  of  which  will  be  in  the  axis  of  x  at  a 

2x 
distance  —  from  0  (Art  78,  Ex.  1,  Cor.),     Hence, 


Fig.  44 


f*2a  2x  , 
f      —%ynx  ax 


I     %y  ttx  dx 


/     x2  <\/2ax  —  x2  dx 
A  Jo 

77      /l2«        '   , ZZZT 

/     x  V^ax  —  x2  dx 
Jo 


5a 

5? 


2.  Find  the  centre  of  gravity  of  a  right  pyramid  of  uni- 
lorm  density,  whose  base  is  any  regular  plane  figure. 

Let  the  vertex  of  the  pyramid  be  the  origin,  and  the  axis 
of  the  pyramid  the  axis  of  x;  divide  the  pyramid  into  slices 
of  the  thickness  dx  by  planes  perpendicular  to  the  axis. 
Then  as  the  areas  of  these  sections  are  as  the  squares  of 
their  homologous  sides,  and  as  the  sides  are  as  their  dis- 
tances from  the  vertex,  so  will  the  areas  of  the  sections  be  as 
the  squares  of  their  distances  from  the  vertex,  and  therefore 
the  masses  of  the  slices  are  as  the  squares  of  their  distances 
from  the  vertex.  Now  imagine  each  slice  to  be  condensed 
into  its  centre  of  gravity,  which  point  is  on  the  axis  of  x. 
Then  the  problem  is  reduced  to  finding  the  centre  of  grav- 


138  THEOREMS   OF  PAPPUS, 

ity  of  a  material  line  in  which  the  density  varies  as  the 
square  of  the  distance  from  one  end,  and  which  may  be 
found  as  in  Ex.  6,  (Art.  78).  Calling  a  the  altitude  of  the 
pyramid,  we  have 


j: 


Xs  dx 


•  =  ^—  =  ¥, 

x*dx 
o 


which  is  the  same  as  in  Art.  75. 

90.  Theorems  of  Pappus.* — (1)  If  a  plane  curve 
revolve  round  any  axis  in  its  plane,  the  area  of  the 
surface  generated  is  equal  to  the  length  of  the 
revolving  curve  multiplied  by  the  length  of  the 
path  described  by  its  centre  of  gravity. 

Let  s  denote  the  length  of  the  curve,  x,  y,  the  co-ordinates 
of  one  of  its  points,  x,  y,  the  co-ordinates  of  the  centre  of 
gravity  of  the  curve;  then,  if  the  curve  is  of  constant 
thickness  and  density,  we  have  from  (2)  of  Art.  78, 


J  yds 


fds 

.•.    2rrys  =  27Tjyds;  (1) 

the  second  member  of  which  is  the  area  of  the  surface 
generated  by  the  revolution  of  the  curve  whose  length  is  s 
about  the  axis  of  x,  (CaL,  Art.  193)  ;  and  the  first  member 
is  the  length  of  the  revolving  curve,  s,  multiplied  by  the 
length  of  the  path  described  by  its  centre  of  gravity,  2ny. 

*  Usually  called  Guldin's  Theorems,  but  originally  enunciated  by  Pappus.    (See 
Walton's  Mechanical  Problems,  p.  42,  3d  Ed.) 


THEOREMS    OF  PAPPUS.  139 

(2)  If  a  plane  area  revolve  round  a  it  //  axis  /'//  its 
plane,  the  volume  generated  is  equal  to  the  area  of 
the  revolving  figure  multiplied  by  the  length  of  the 
path  described  by  its  centre  of  gravity. 

Let  A  denote  the  plane  area,  and  let  it  be  of  constant 
thickness  and  density,  then  (2)  of  Art.  82  becomes 

J  JV  dx  dy 
J  j  dxdy 

or  2ny  I    I  dk  =  2n  /       y  dx  dy, 

(substituting  dA  for  dx  dy), 

.-.    2nyA  =  n  fy2dx,  (2) 

the  integral  being  taken  for  every  point  in  the  perimeter  of 
the  area ;  but  the  second  member  is  the  volume  of  the 
solid  generated  by  the  revolution  of  the  area  (Cal.,  Art. 
203) ;  and  the  first  member  is  the  area  of  the  revolving 
figure,  A,  multiplied  by  the  length  of  the  path  described 
by  its  centre  of  gravity,  2ny. 

Cor. — If  the  curve  or  area  revolve  through  any  angle,  6, 
instead  of  2rr,  (1)  and  (2)  become 


6 ys  —  6  I yds, 

(3) 

and                             OyA  =  \d  f  y*  dx, 

(4) 

and  the  theorems  are  still  true. 

Sch. — If  the  axis  cuts  the  revolving  curve  or  area,  the 
theorems  still  apply  with  the  convention  that  the  surface 
or  volume  generated  by  the  portions  of  the  curve  or  area  on 
opposite  sides  of  the  axis  are  affected  with  opposite  signs. 


140  EXAMPLES. 


EXAMPLES. 

1.  A  circle  of  radius,  a,  revolves  round  an  axis  in  its  own 
plane  at  a  distance,  c,  from  its  centre;  find  the  surface  of 
the  ring  generated  by  it. 

The  length  (circumference)  of  the  revolving  curve  = 
2tt<7,  ;  the  length  of  the  path  described  by  its  centre  of 
gravity  =  2ttc; 

.  * .     the  area  of  the  surface  of  the  ring  =  4zn2ac. 

2.  An  ellipse  revolves  round  an  axis  in  its  own  plane, 
the  perpendicular  distance  of  which  from  the  centre  is  c  ; 
find  the  volume  of  the  ring  generated  during  a  complete 
revolution. 

Let  a  and  b  be  the  semi-axes  of  the  ellipse ;  then  the 
revolving  area  =  nab',  the  length  of  the  path  described  by 
its  centre  of  gravity  =  2ttc  ; 

.*.    the  volume  of  the  ring  =  2n2abc. 

Observe  that  the  volume  is  the  same  for  any  position  of  the  axes 
of  the  ellipse  with  respect  to  the  axis  of  revolution,  provided  the  per- 
pendicular distance  from  that  axis  to  the  centre  of  the  ellipse  is  the 
same. 

3.  The  surface  of  a  sphere,  of  radius  aa  =  4rr«2;  the 
length  of  a  semi-circumference  =  net  ;  find  the  length  of 
the  ordinate  to  the  centre  of  gravity  of  the  arc  of  a  semi- 
circle. .        _       2a 

An*,  y  =  — 

7T 

4.  The   volume  of  a  sphere,  of  radius  a,   =  ^rras ;   the 

area  of  a  semicircle  =  i^a2;  find  the  distance  of  the  centre 

of  gravity  of  the  semicircle  from  the  diameter. 

4         -         4*i 
Ans.  y  =  — . 

OH" 

5.  A  circular  tower,  the  diameter  of  which  is  20  ft.,  is 
being  built,  and  for  every  foot  it  rises  it  inclines  1  in.  from 


EXAMPLES.  141 

the  vertical ;  find  the  greatest  height  it  can  reach  vvirrmut 
falling.  Am.   2  lu  It. 

G.  A  circular  table  weighs  20  lbs.  and  rests  on  four  legs 
in  its  circumference  forming  a  square;   find  the  least   ver- 
tical pressure  that  must  be  applied  at  its  edge  to  overturn  it. 
Ans.  20_(\/2  +  1)  =  48.28.11ks. 

7.  If  the  sides  of  a  triangle  be  3,  4,  and  5  feet,  find  the 
distance  of  the  centre  of  gravity  from  each  side. 

Ans.  4,  1,  |  ft. 

8.  An  equilateral  triangle  stands  vertically  on  a  rough 
plane  ;  find  the  ratio  of  the  height  to  the  base  of  the  plane 
when  the  triangle  is  on  the  point  of  overturning. 

Ans.   \/3  :  1. 

9.  A  heavy  bar  14  feet  long  is  bent  into  a  right  angle  so 
that  the  lengths  of  the  portions  which  meet  at  the  angle 
are  8  feet  and  6  feet  respectively ;  show  that  the  distance 
of  the  centre  of  gravity  of  the  bar  so  bent  from  the  point 
of  the  bar  which  was  the  centre  of  gravity  when  the  bar 

was  straight,  is  — - —  feet. 

10.  An  equilateral  triangle  rests  on  a  square,  and  the  base 
of  the  triangle  is  equal  to  a  side  of  the  square  ;  find  the 
centre  of  gravity  of  the  figure  thus  formed. 

Ans.  At  a  distance  from  the  base  of  the  triangle  equal  to 
3 

■ —  of  the  base. 

8  f  2\/3 

11.  Find  the  inclination  of  a  rough  plane  on  which  half 
a  regular  hexagon  can  just  rest  in  a  vertical  position  with- 
out overturning,  with  the  shorter  of  its  parallel  sides  in 
contact  with  the  plane.  Ans.  3  a/3  *  5- 

12.  A  cylinder,  the  diameter  of  which  is  10  ft,  and  height 
60  ft.,  rests  on  another  cylinder  the  diameter  of  which  is 


142  EXAMPLES. 

18  ft.,  and  height  6  ft. ;  and  their  axes  coincide  ;  find  then 
common  centre  of  gravity.  Ans.  27§|J  ft.  from  the  base. 

13.  Into  a  hollow  cylindrical  vessel  11  ins.  high,  and 
weighing  10  lbs.,  the  centre  of  gravity  of  which  is  5  ins. 
from  the  base,  a  uniform  solid  cylinder  6  ins.  long  and 
weighing  20  lbs.,  is  just  fitted  ;  find  their  common  centre  of 
gravity.  Ans.  3f  ins.  from  base. 

14.  The  middle  points  of  two  adjacent  sides  of  a  square 
are  joined  and  the  triangle  formed  by  this  straight  line  and 
the  edges  is  cut  off;  find  the  centre  of  gravity  of  the 
remainder  of  the  square. 

Ans.  -£r  of  diagonal  from  centre. 

15.  A  trapezoid,  whose  parallel  sides  are  4  and  12  ft. 
Jong,  and  the  other  sides  each  equal  to  5  ft.,  is  placed  with 
its  plane  vertical,  and  with  its  shortest  side  on  an  inclined 
plane ;  find  the  relation  between  the  height  and  base  of  the 
plane  when  the  trapezoid  is  on  the  point  of  falling  over. 

A?is.  8  :  7. 

16.  A  regular  hexagonal  prism  is  placed  on  an  inclined 
plane  with  its  end  faces  vertical ;  find  the  inclination  of 
the  plane  so  that  the  prism  may  just  tumble  down  the  plane. 

Ans.  30°. 

17.  A  regular  polygon  just  tumbles  down  an  inclined 
plane  whose  inclination  is  10° ;  how  many  sides  has  the 
polygon  ?  Ans.  18. 

18.  From  a  sphere  of  radius  R  is  removed  a  sphere  of 
radius  r,  the  distance  between  their  centres  being  c ;  find 
the  centre  of  gravity  of  the  remainder. 

Ans.  It  is  on  the  line  joining  their  centres,  and  at  a  dis- 

c? 

tance  ^— from  the  centre. 

Bs  —  r3 

19.  A  rod  of  uniform  thickness  is  made  up  of  equal 
lengths  of  three  substances,  the  densities  of  which  taken  in 


EXAMPLES.  143 

order  are  in  the  proportion  of  1,  2,  and  3  ;  find  the  position 

of  the  centre  of  gravity  of  the  rod. 

Ans.  At  i\  of  the  whole  length  from  the  end  of  the 
densest  part. 

20.  A  heavy  triangle  is  to  be  suspended  by  a  string  pass- 
ing through  a  point  on  one  side ;  determine  the  position  of 
the  point  so  that  the  triangle  may  rest  with  one  side 
vertical. 

Ans.  The  distance  of  the  point  from  one  end  of  the  side 
=  twice  its  distance  from  the  other  end. 

21.  The  sides  of  a  heavy  triangle  are  3,  4,  5,  respectively ; 
if  it  be  suspended  from  the  centre  of  the  inscribed  circle 
show  that  it  will  rest  with  the  shortest  side  horizontal. 

22.  The  altitude  of  a  right  cone  is  h,  and  a  diameter  of 
the  base  is  b ;  a  string  is  fastened  to  the  vertex  and  to  a 
point  on  the  circumference  of  the  circular  base,  and  is  then 
put  over  a  smooth  peg ;  show  that  if  the  cone  rests  with  its 
axis  horizontal  the  length  of  the  string  is  V(h2  +  &). 

23.  Find  the  centre  of  gravity  of  the  helix  whose  equa- 
tions are 

x  =  a  cos  <f> ;     y  =  a  sin  <p  ;     z  =  1ca<p. 

Ans.  x  =  ha~\  y  =  ha 
z 

24.  Find  the  distance  of  the  centre  of  gravity  of  the 
catenary  (Cal.,  Art.  177),  from  the  axis  of  x,  the  curve 
being  divided  into  two  equal  portions  by  the  axis  of  y. 

Ans.  If  2?  is  the  length  of  the  curve  and  (h,  h)  is  the 
extremity,  the  centre  of  gravity  is  on  the  axis  of  y  at  a 

distance  — ^ —  from  the  axis  of  x. 


144  EXAMPLES. 

25.  Find  the  centre  of  gravity  of  the  area  included 
between  the  arc  of  the  parabola,  if  =  ±ax,  and  the  straight 
line  y  =  Jcx.  .        _        8a  2a 

26.  Find  the  centre  of  gravity  of  the  area  bounded  by 
the  cissoid  and  its  asymptote,  the  equation  of  the  cissoid 

Xs 

being  y2  == Ans.  x  —  %a. 

&a  —  x 

27.  Find  the  centre  of  gravity  of  the  area  of  the  witch 
of  Agnesi. 

Ans.  At  a  distance  from  the  asymptote  equal  to  |  of  the 
diameter  of  the  base  circle. 

28.  Find  the  centre  of  gravity  of  the  area  included  be- 
tween the  arc  of  a  semi-cycloid,  the  circumference  of  the 
generating  circle,  and  the  base  of  the  cycloid,  the  common 
tangent  to  the  circle  and  cycloid  at  the  vertex  of  the  latter 
being  taken  as  axis  of  x,  the  vertex  being  origin,  and.  a  the 
radius  of  the  generating  circle. 

Ans.  x  =  — ■ a:  y  =  fa. 

4r7T  4 

29.  Find  the  centre  of  gravity  of  the  area  contained  be- 
tween   the   curves   y2  =  ax  and   y2  =  2ax  —  x2,  which  is 

above  the  axis  of  x.  ■    .        _           15  n  —  44    _  a 

Ans.  x  =  a—- — ;  y 


157r_407  '        Stt  —  8 

30.  Find  the  centre  of  gravity  of  the  area  included  by 
the  curves  y2  =  ax  and  x2  =  ly. 

Ans.  x  =  -io^hf ;  y  =  ^9-o«f^. 

31.  Find  the  distance  of  the  centre  of  gravity  of  the  area 
of  the  circular  sector,  BOCA,  (Fig.  39),  from  the  centre. 

Let  20  =  the  angle  included  by  the  bounding  radii. 

Ans.  x  =  f a  —^ — 


EXAMPLES.  145 

32.  Find  the  distance  of  the  centre  of  gravity  of  the 
circular  segment,  BCA,  (Fig.  39),  from  the  centre. 

a  sin3  6  BC3 

1    V  —  siu  0cos0  —  12  area  of  ABC' 

33.  Find  the  centre  of  gravity  of  the  area  bounded  by 
the  cardioid  r  =  a{\  -\-  cos  0).  Ans.  x  =  fa. 

34.  Find  the  centre  of  gravity  of  the  area  included  by  a 

loop  of  the  curve  r  =  a  oos  20.  128«  y/% 

Ans.  x  =  — — 

1057T 

35.  Find  the  centre  of  gravity  of  the  area  included  by  a 
loop  of  the  curve  r  =  a  cos  39.  gla  ^3 

An8-'x  =  -mV' 

36.  Find  the  centre  of  gravity  of  the  area  of  the 
sector  in  Eex.  31,  if  the  density  varies  directly  as  the  dis- 
tance from  the  centre.  .        .        3a    sin  6 

Ans.  x  =3  -j-  •  — - — 

4      e 

37.  Find  the  centre  of  gravity  of  the  area  of  a  circular 
sector  in  which  the  density  varies  as  the  nth  power  of  the 
distance  from  the  centre. 

Ans.   =  •  -r,  where  a  is  the  radius  of  the  circle,  I  the 

n  +  3      I  ' 

length  of  the  arc,  and  c  the  length  of  the  chord,  of  the 

sector. 

38.  Find  the  centre  of  gravity  of  the  area  of  a  circle  in 
which  the  density  at  any  point  varies  as  the  nth  power  of 
the  distance  from  a  given  point  on  the  circumference. 

Ans.  It  is  on  the  diameter  passing  through  the  given 

2  in  4-  2) 
point  at  a  distance  from  this  point  equal  to  a% 

a  being  the  radius* 
7 


146  EXAMPLES. 

39.  Find  the  centre  of  gravity  of  the  area  of  a  quadrant 
of  an  ellipse  in  which  the  density  at  any  point  varies  as 
the  distance  of  the  point  from  the  major  axis. 

Ans.  x  =  la-,y=—b. 

40.  Find  the  distance  of  the  centre  of  gravity  of  the  sur- 
face of  a  cone  from  the  vertex. 

Let  a  =  the  altitude.  Ans.  x  =  \a. 

41.  Find  the  centre  of  gravity  of  the  surface  formed  by 
revolving  the  curve 

r  =  a  (1  +  cos  0), 

round  the  initial  line.  Ans.  x  =  — --. 

42.  A  parabola  revolves  round  its  axis ;  find  the  centre 
of  gravity  of  a  portion  of  the  surface  between  the  vertex 
and  a  plane  perpendicular  to  the  axis  at  a  distance  from 
the  vertex  equal  to  f  of  the  latus  rectum. 

Ans.  Its  distance  from  the  vertex  =  •§-$-  (latus  rectum). 

43.  Find  the  centre  of  gravity  of  a  cone,  the  density  of 
each  circular  slice  of  which  varies  as  the  nth  power  of  its 
distance  from  a  parallel  plane  through  the  vertex. 

Let  the  vertex  be  the  origin  and  a  the  altitude. 

A        -       n  +  3 

Ans.  x  = a. 

n  +  4 

44.  Find  the  centre  of  gravity  of  a  cone,  the  density  of 
every  particle  of  which  increases  as  its  distance  from  the 
axis. 

Ans.  x  =  \a,  where  the  vertex  is  the  origin  and  a  the 
altitude. 

45.  Find  the  centre  of  gravity  of  the  volume  of  uniform 
density  contained  between  a  hemisphere  and  a  cone  whose 
vertex  is  the  vertex  of  the  hemisphere  and  base  is  the  base 
of  the  hemisphere. 


EXAMPLES.  147 

Ans.  x  =  -,  where  the  vertex  is  the  origin  and  a  the 
altitude. 

46.  Find  the  distance  of  the  centre  of  gravity  of  a  hemi- 
sphere from  the  centre,  the  radius  being  a. 

Ans.   x  =  |«. 

47.  Find  the  centre  of  gravity  of  the  solid  generated  by 

the  revolution  of  the  semicycloid, 

/ x 

y  =  y'Zax  —  x2  -f  a  vers-1  - , 

(1)  round  the  axis  of  x,  and  (2)  round  the  axis  of  y. 

Ans.   (1)  *  _    g  ^  _  l6)  ,    (^  y  _  ^y  +  -^  _. 

48.  Find  the  centre  of  gravity  of  the  volume  formed  by 
the  revolution  round  the  axis  of  x  of  the  area  of  the  curve 

y*  —  axy2  -f  #*  =  0. 

Ans.  x  =  -— • 

49.  Find  the  centre  of  gravity  of  the  volume  generated 

by  the  revolution  of  the  area  in  Ex.  29  round  the  axis  of  y. 

.        _  ha 

Ans.  y  =  cT7— t-7t' 

2  (Iott  —  44) 

50.  Find  the  centre  of  gravity  of   a   hemisphere  when 

the  density  varies  as  the  square  of  the  distance  from  the 

centre.  .        -        oa 

Ans.  x  =  — • 
\& 

51.  Find  the  centre  of  gravity  of  the  solid  generated  by  a 
semi-parabola  bounded  by  the  latus  rectum,  revolving 
round  the  latus  rectum. 

Ans.  Distance  from  focus  =  -^  of  latus  rectum. 


148  EXAMPLES. 

52.  The  vertex  of  a  right  circular  cone  is  at  the  centre  of 
a  sphere  ;  find  the  centre  of  gravity  of  a  body  of  uniform 
density  contained  within  the  cone  and  the  sphere. 

Ans.  The  distance  of  the  centre  of  gravity  from  the  ver- 
tex of  the  cone  =  —  (1  +  cos  «),  where  a  =  the  semi- 

o 

vertical   angle   of   the   cone   and  a  =  the  radius  of   the 
sphere. 

53.  Find  the  distance  from  the  origin  to  the  centre  of 
gravity  of  the  solid  generated  by  the  revolution  of  the 
cardioid  round  its  prime  radius,  its  equation  being 

r  =  a  (1  -J-  cos  0). 

Ans.  x  =  \a. 

54.  Find  by  Art.  90  (1)  the  surface  and  (2)  the  volume 
of  the  solid  formed  by  the  revolution  of  a  cycloid  round 
the  tangent  at  its  vertex. 

Ans.  Surface  =  -^-7ra2;  Volume  =  7r2a3. 

55.  Find  (1)  the  surface  and  (2)  the  volume  of  the  solid 
formed  by  the  revolution  of  a  cycloid  round  its  base. 

Ans.   (1)  6/™2 ;  (2)  5nW. 

56.  An  equilateral  triangle  revolves  round  its  base, 
whose  length  is  a ;  find  (1)  the  area  of  the  surface, 
and  (2)  the  volume  of  the  figure  described. 


Ans.  (1)  t™2  V3  ;  (2)  ™ 


57.  Find  (1)  the  surface  and  (2)  the  volume  of  a  ring 
with  a  circular  section  whose  internal  diameter  is  12  ins., 
and  thickness  3  ins. 

Ans.  (1)  444.1  sq.  in.;  (2)  333.1  cub.  in. 


CHAPTER    V. 

FRICTION. 

91.  Friction.  —Friction  is  that  force  which  acts  between 
t'vo  bodies  at  their  surface  of  contact,  and  in  the  direction 
of  a  tangent  to  that  surface,  so  as  to  resist  their  sliding  on 
each  other.  .  It  depends  on  the  force  with  which  the  bodies 
are  pressed  together.  All  the  curves  and  surfaces  which  we 
have  hitherto  considered  were  supposed  to  be  smooth,  and, 
as  such,  to  offer  no  resistance  to  the  motion  of  a  body  in 
contact  with  them  in  any  other  than  a  Dormal  direction. 
Such  curves  and  surfaces,  however,  are  not  to  be  found  in 
nature.  Every  surface  is  capable  of  destroying  a  certain 
amouut  of  force  in  its  tangent  plane,  i.e.,  it  possesses  a  certain 
degree  of  roughness,  in  virtue  of  which  it  resists  the  sliding 
of  other  surfaces  upon  it.  This  resistance  is  called  friction, 
and  is  of  two  kinds,  viz.,  sliding  and  rolling  friction.  The 
first  is  that  of  a  heavy  body  dragged  on  a  plane  or  other 
surface,  an  axle  turning  in  >a  fixed  box,  or  a  vertical  shaft 
turning  on  a  horizontal  plate.  Friction  of  the  second  kind 
is  that  of  a  wheel  rolling  along  a  plane.  Both  kinds  of 
friction  are  governed  by  the  same  laws;  the  former  is  much 
greater  than  the  latter  under  the  same  circumstances,  and 
is  the  only  one  that  we  shall  consider. 

A  smooth  surface  is  one  which  opposes  no  resistance  to 
the  motion  of  a  body  upon  it.  A  rough  surface  is  one 
which  does  oppose  a  resistance  to  the  motion  of  a  body 
upon  it. 

The  surfaces  of  all  bodies  consist  of  very  small  elevations  and 
depressions,  so  that  if  they  are  pressed  against  each  other,  the 
elevations  of  one  fit.  more  or  less,  into  the  depressions  of  the  other, 
and  the  surfaces  interpenetrate  each  other ;  and  the  mutual  penetra- 


150  LAWS   OF  FRICTION. 

tion  is  of  course  greater,  if  the  pressing  force  is  greater.  Hence, 
when  a  force  is  applied  so  as  to  cause  one  body  to  move  on  another 
with  which  it  is  in  contact,  it  is  necessary,  before  motion  can  take 
place,  either  to  break  off  the  elevations  or  compress  them,  or  force  the 
'bodies  to  separate  far  enough  to  allow  them  to  pass  each  other. 
Much  of  this  roughness  may  be  removed  by  polishing;  and  the  effect 
of  much  of  it  may  be  destroyed  by  lubrication. 

Friction  always  acts  along  a  tangent  to  the  surface  at  the  point  of 
contact ;  and  its  direction  is  opposite  to  that  of  the  line  of  motion  ;  it 
presents  itself  in  the  motion  of  a  body  as  a  passive  force  or  resistance* 
since  it  can  only  hinder  motion,  but  can  never  produce  or  aid  it.  In 
investigations  in  mechanics  it  can  be  considered  as  a  force  acting  in 
opposition  to  every  motion  whose  direction  lies  in  the  plane  of  contact 
of  the  two  bodies.  Whatever  may  be  the  direction  in  which  we  move 
a  body  resting  upon  a  horizontal  or  inclined  plane,  the  friction  will 
always  act  in  the  opposite  direction  to  that  of  the  motion,  *.  e.,  when 
we  slide  a  body  down  an  inclined  plane,  it  will  appear  as  a  force  up 
the  plane.  A  surface  may  also  resist  sliding  motion  by  means  of  th3 
adhesion  between  its  substance  and  that  of  another  body  in  contact 
with  it.f 

The  friction  of  a  body  on  a  surface  is  measured  by  the 
least  force  which  will  put  the  body  in  motion  along  the 
surface. 

92.  Laws  of  Friction. — In  our  ignorance  of  the 
constitution  of  bodies,  the  laws  of  friction  must  be  deduced 
from  experiment.  Experiments  made  by  Coulomb  and 
Morin  have  established  the  following  laws  of  friction : 

(1)  The  friction  varies  as  the  normal  pressure  when  the 
materials  of  the  surfaces  in  contact  remain  the  same.  Subse- 
quent experiments  have,  however,  considerably  modified 
this  law,  and  shown  that  it  can  be  regarded  only  as  an 
approximation  to  the  truth.  When  the  pressure  is  very 
great  it  is  found  that  the  friction  is  less  than  this  law 
would  give. 

*  Weisbach,  p.  309. 

t  See  Rankiue.'s  Applied  Mechanics,  p.  209. 


LAWS   OF  FRICTIOX.  151 

(2)  The  friction  is  independent  of  the  extent  of  the  sur- 
faces in  contact  so  long  as  the  normal  pressure  remains  the 
same.  When  the  surfaces  in  contact  are  very  small,  as  for 
instance  a  cylinder  resting  on  a  surface,  this  law  gives  the 
friction  much  too  great. 

These  two  laws  are  true  when  the  body  is  on  the  point  of  moving, 
and  also  when  it  is  actually  in  motion ;  but  in  the  case  of  motion  the 
magnitude  of  the  friction  is  not  always  the  same  as  when  the  body  is 
beginning  to  move  ;  when  there  is  a  difference,  the  friction  is  greater 
in  the  state  bordering  on  motion  than  in  actual  motion. 

(3)  The  friction  is  independent  of  the  velocity  when  the 
tody  is  in  motion. 

It  follows  from  these  laws  that,  if  R  be  the  normal 
pressure  between  the  bodies,  F  the  force  of  friction,  and  u 
the  constant  ratio  of  the  latter  to  the  former  when  slipping 
is  about  to  ensue,  we  have 

F  =  iiR.  (1) 

The  fraction  \i  is  called  the  co-efficient  of  friction  ;  and  if 
the  first  law  were  true,  \i  would  be  strictly  constant  for  the 
same  pair  of  bodies,  whatever  the  magnitude  of  the  normal 
pressure  between  them  might  be.  This,  however,  is  not 
the  case.  When  the  normal  pressure  is  nearly  equal  to  that 
which  would  crush  either  of  the  surfaces  in  contact,  the 
force  of  friction  increases  more  rapidly  than  the  normal 
pressure.  Equation  (1)  is  nevertheless  very  nearly  true 
when  the  differences  of  normal  pressure  are  not  very  great ; 
and  in  what  follows  we  shall  assume  this  to  be  the  case. 

Remark. — The  laws  of  friction  were  established  by  Coulomb,  a 
distinguished  French  officer  of  Engineers,  and  were  founded  on 
experiments  made  by  him  at  Rochefort.  The  results  of  these  experi- 
ments were  presented  in  1781  to  the  French  Academy  of  Sciences,  and 
in  1785  his  Memoir  on  Friction  was  published.  A  very  full  abstract 
of  this  paper  is  given  in  De  Young's  Natural  Philosophy ,  Vol.  II, 
p.  170  (1st  Ed.).  Further  experiments  were  made  at  Metz  by  Morin, 
1831-34,  by  direction  of  the  French  military  authorities,  the  result  of 


152  ANGLE   OF  FRICTION. 

which  has  been  to  confirm,  with  slight  exceptions,  all  the  results  oi 
Coilomb,  and  to  determine  wilh  considerable  precision  the  numerical 
values  of  the  coefficients  of  friction,  for  all  the  substances  usually 
employed  in  the  construction  of  machines.  (See  Galbraith's  Me- 
chanics, p.  68,  Twisden's  Practical  Mechanics,  p.  138,  and  Weisbach's 
Mechanics,  Vol.  I,  p.  317.) 

93.  Magnitudes  of  Coefficients  of  Friction. — Prac- 
tically there  is  no  observed  coefficient  much  greater  than  1. 
Most  of  the  ordinary  coefficients  are  less  than  J.  The  fol- 
lowing results,  selected  from  a  table  of  coefficients,*  will 
afford  an  idea  of  the  amount  of  friction  as  determined  by 
experiment ;  these  results  apply  to  the  friction  of  motion. 

For  iron  on  stone      \i  varies  between  .3  and  .7. 
For  timber  on  timber "       "  "        .2  and  .5. 

For  timber  on  metals  "       "  "        .2  and  .6. 

For  metals  on  metals  "*     "  "      .15  and  .25. 

For  full  particulars  on  this  subject  the  student  is  referred 
to  Rankine's  Applied  Mechanics,  p.  209,  and  Moseley's 
Engineering,  p.  124,  also  to  the  treatise  of  M.  Morin,  where 
he  will  find  the  subject  investigated  in  all  its  completeness. 

94.  Angle  of  Friction.— The  angle  at  which  a  rough 
'plane  or  surface  may  be  inclined  so  that  a  body,  when  acted 
upon  by  the  force  of  gravity  only, may  just  rest  upon  it  ivith- 
out  sliding,  is  called  the  Angle  of  Friction.\ 

Let  «  be  the  angle  of  inclination  of 
the  plane  AB  just  as  the  weight  is  on 
the  point  of  slipping  down;  W  the 
weight  of  the  body  ;  E  the  normal  pres- 
sure on  the  plane ;  F  the  force  of  fric-  Fig.45 
tion  acting  along  the  plane  =  fiR  (Art. 
92).  Then,  resolving  the  forces  along  and  perpendicular  to 
the  plane  we  have  for  equilibrium 

*  Rankine's  Applied  Mechanics,  p.  211. 

t  Sometimes  called  "the  angle  of  repose;'"'  aiso  called  "the  limiting  angle  ol 
resistance." 


REACTION   OF  A    ROUGH   CURVE.  153 

uR  =  W  sin  «  ;  R  =  H '  cos  a ; 

.-,     tan  «  ::-  [i.,  (1) 

which  gives  the  limiting  value  of  the  inclination  of  the 
plane  for  which  equilibrium  is  possible.  The  body  will  rest 
on  the  plane  when  the  angle  of  inclination  is  less  than  the 
angle  of  friction,  and  will  slide  if  the  angle  of  inclination 
exceeds  that  angle;  and  this  will  be  the  case  however  great 
W  may  be;  the  reason  being  that  in  whatever  manner 
we  increase  W,  in  the  same  proportion  we  increase  the 
friction  upon  the  plane,  which  serves  to  prevent  W from 
sliding. 

From  (1)  we  see  that  the  tangent  of  the  angle  of  friction 
is  equal  to  the  coefficient  of  friction. 

95.  Reaction  of  a  Rough  Curve 
or  Surface. — Let  AB  be  a  rough  curve 
or  surface ;  P  the  position  of  a  particle 
ou  it ;  and  suppose  the  forces  acting  on 
P  to  be  confined  to  the  plane  of  the 
paper.  Let  Rx  =  the  normal  resistance  of  the  surface, 
acting  in  the  normal,  PN,  and  F  =  the  force  of  friction, 
acting  along  the  tangent,  PT. 

The  resultant  of  Rx  and  F,  called  the  Total  Resistance* 
of  the  surface,  is  represented  in  magnitude  and  direction  by 
the  line  PR  =  R,  which  is  the  diagonal  of  the  parallelo- 
gram determined  by  Rx  and  F.  We  have  seen  that  the 
total  resistance  of  a  smooth  surface  is  normal  (Art.  41)  ;  but 
this  limitation  does  not  apply  to  a  rough  surface.  Let  (ft 
denote  the  angle  between  R  and  the  normal  Rt  ;  then  <j>  is 
given  by  the  equation 

F 

tan  0  =  77— 
R\ 

*  Minchin's  Statics,  p.  54. 


154  FRICTION   ON  AN  INCLINED  PLANE, 

Hence,  <j>  will  be  a  maximum  when  the  force  of  friction, 
F,  bears  the  greatest  ratio  to  the  normal  pressure  Rv  But 
this  greatest  ratio  is  attained  when  the  body  is  just  on  the 
point  of  slipping  along  the  surface,  and  is  what  we  called 
the  coefficient  of  friction  (Art.  92),  that  is 

F 

R[  =  ^ 

.*.    tan  (f>  =  p. 

Therefore  the  greatest  angle  by  which  the  Total  Resistance 
of  a  rough  curve  or  surface  can  deviate  from  the  normal  is 
the  angle  whose  tangent  is  the  coefficient  of  friction  for  the 
bodies  in  contact ;  and  this  deviation  is  attained  when  slip- 
ping is  about  to  commence. 

Coe.— By  (1)  of  Art.  94,  tan  «  =  ft; 

.-.    0  =  «; 

hence,  the  direction  of  the  total  resistance,  R,  is  inclined  at 
an  angle  a  to  the  normal ;  i.  e.,  the  greatest  angle  that  the 
Total  Resistance  of  a  rough  curve  or  surface  can  make  with 
the  normal  is  equal  to  the  angle  of  friction,  corresponding 
to  the  two  bodies  in  contact. 

96.  Friction  on  an  Inclined  Plane. — A  body  rests  on 
a  rough  inclined  plane,  and  is  acted  on  by  a  given  force,  P, 
in  a  vertical  plane  which  is  perpendicular  to  the  inclined 
plane  ;  find  the  limits  of  the  force,  and  the  angle  at  which 
the  least  force  capable  of  drawing  the  particle  up  the  plane 
must  act. 

Let  i  =  the  inclination  of  the  plane  to  the  horizon ;  0  = 
the  angle  between  the  inclined  plane  and  the  line  of  action 
of  P;  \i  =  the  coefficient  of  friction;  and  let  us  first  sup- 
pose that  the  body  is  on  the  point  of  moving  down  the 


FRICTION  ON  AN  INCLINED   PLANE.  155 

plane,  so  that  friction  is  a  force  acting  up  the  plane,  then 
resolving  along,  and  perpendicular  to,  the  plane,  we  have 

F  +  Pcosd  =  Jf  sin  t, 

8  +  Psind  =  Wcosi, 

F  =  [iR; 

...    p=w^i-^cosi 

cos  0  —  \i  sin  9  x  ' 

And  if  P  is  increased  so  that  motion  vp  the  plane  is  just 
beginning,  F  acts  in  an  opposite  direction,  and  therefore 
the  sign  of  ^  must  be  changed  and  we  have 

p__  yrini  +  ftCQBi^ 

cos  0  +  ^  sin  0  *  ' 

Hence,  there  will  be  equilibrium  if  the  body  be  acted  on  by 
a  force,  the  magnitude  of  which  lies  between  the  values  of 
P  in  (1)  and  (2).  Substituting  tan  0  for  \i  (Art.  95) ;  (2) 
becomes 

P=r?M_+*L  (3) 

cos  (0  —  0)  ' 

To  determine  6  in  (2)  so  that  P  shall  be  a  minimum  we 
must  put  the  first  derivative  of  P  with  respect  to  0  =  0, 
therefore 

dP       ro-  /  .     .  ,  -x     sin  B  —  fj,  cos  0 

-j-  =  Tf  (sin  *  +  jw  cos  i)  t 3-,        . — Hxo  =  °  > 

.  • .    tan  0  =  ^ ; 

that  is,  the  force  P  necessary  to  draw  the  body  up  the  plane 
will  be  the  least  possible  when  0  =  the  angle  of  friction. 


156  DOUBLE-INCLINED  PLANE. 

Hence  we  infer  that  a  given  force  acts  to  the  greatest 
advantage  in  dragging  a  weight  up  a  hill,  if  the  angle  at 
which  its  line  of  action  is  inclined  to  the  hill  is  equal  to 
the  angle  of  friction  of  the  hill.  Similarly,  a  force  acts  to 
the  greatest  advantage  in  dragging  a  weight  along  a  hori- 
zontal plane  if  its  line  of  action  is  inclined  to  the  plane 
at  the  angle  of  friction  of  the  plane.  We  may  also  deter- 
mine from  this  the  angle  at  which  the  traces  of  a  drawing 
horse  should  be  inclined  to  the  plane  of  traction. 

These  results  are  those  which  are  to  be  expected,  because 
some  part  of  the  force  ought  to  be  expended  in  lifting  the 
weight  from  the  plane,  so  that  friction  may  be  diminished. 
(See  Price's  Anal.  Mech's,  Vol.  I,  p.  160.) 

97.   Friction  on    a    Double-Inclined  Plane. — Two 

bodies,  whose  weights  are  P  and  Q,  rest  on  a  rough  double- 
inclined  plane,  and  are  connected  by  a  string  which  passes 
over  a  smooth  peg  at  a  point,  A,  vertically  over  the  intersec- 
tion, B,  of  the  two  planes.  Find  the  position  of  equili- 
brium. 

Let  a  and  j3  be  the  inclinations  of 
the  two  planes  ;  let  I  =  the  length  of 
the  string,  and  h  =  AB;  and  let  0 
and  0'  be  the  angles  the  portions  of 
tnt  string  make  with  the  planes. 

Suppose  P  is  on  the  point  of 
ascending,  and  Q  of  descending. 
Then,  since  the  motion  of  each  body  is  about  to  ensue,  the 
total  resistances,  R  and  S,  must  each  make  the  angle  of 
friction  with  the  corresponding  normal  (Art.  95,  Cor.) ;  and 
since  the  weight,  P,  is  about  to  move  upwards  the  friction 
must  act  downwards,  and  therefore  R  must  lie  below  the 
normal,  while,  since  Q  is  about  to  move  dowmwards,  the 
friction  must  act  upwards,  and  therefore  S  must  be  above 
the  normal. 


DOUBLE-INCLINED   PLANE.  157 

If  T  is  the  tonsion  of  the  string,  we  have  for  the  equi- 
librium of  P,  (Art.  32), 

cos  (0  —  0) 

Aud  for  the  equilibrium  of  Q, 

sin  (fl  —  0) 
V  cos  (0'  +  0)' 

Equating  the  values  of  T  we  get 

sin  («  +  0)  _  ^sin  (3-0)  m 

^ cos  ((9-0)  ~~  Vcos(0'  +  0)'  V  ; 

and  if  P  is  about  to  move  down  the  plane,  the  friction  acts 
in  an  opposite  direction,  and  therefore  the  sign  of  0  must 
be  changed  and  we  have 

sin  (a  —  0)  _      sin  (0  +  0)  ,  , 

cos  ((9  +  0)  ~~  V  cos  (d'  -  0)'  v  ; 

(1)  or  (2)  is  the  only  statical  equation  connecting  the 
given  quantities. 

We  obtain  a  geometric  equation  by  expressing  the  length 
of  the  string  in  terms  of  h,  a,  (3,  6,  and  6',  which  is 

/cos  a       cos  0\  .  . 

1  = h  (et  +  arsy"  (3). 

From  (1)  or  (2)  and  (3)  the  values  of  0  and  6'  can  be  found, 
and  this  determines  the  positions  of  P  and  Q. 

Otherwise  thus  : 

Instead  of  considering  the  total  resistances,  R  and  S,  we 
may  consider  two  normal  resistances,  Nt  and  S19  and  two 


158  DOUBLE-INCLINED   PLANE. 

forces  of  friction,  [iRl  and  fiS\,  acting  respectively  down 
the  plane  a  and  up  the  plane  j3.  In  this  case,  considering 
the  equilibrium  of  P,  and  resolving  forces  along,  and  per- 
pendicular to,  the  plane  a,  we  have 

P  sin  a  +  pB-L  =  T  cos  0, )  .  . 

P  cos  a  =  Rx  +  T  sin  (9,    (  W 

and  for  the  equilibrium  of  (), 

<)sin0  =^  +  r  cos  0',) 

Gcos]8  =  ^  +  rsin6>'.    J  {  } 

Eliminating  Rt,  8V  and  T  from  (4)  and  (5)  we  get  (1), 
the  same  statical  equation  as  before. 

The  method  of  considering  total  resistances  instead  of 
their  normal  and  tangential  components  is  usually  more 
simple  than  the  separate  consideration  of  the  latter  forces. 
{See  Minchin's  Statics,  p.  60.) 

Cor. — If  Q  is  given  and  P  be  so  small  that  it  is  about  to 
ascend,  its  value,  Pt,  will  be  given  by  (1), 

_      sin  (0  -  0)  cos  (0  -  0) 

1  ~  V  sin  (a  +  0)  cos  (0'  +  <p)'  W 

and  if  P  is  so  large  that  it  is  about  to  drag  Q  up,  its  value, 
P2,  will  be  given  by  (2) 

_      sin  (fl  +  0)  cos  (0  +  0)  m 

2  -  v  sin  («  -  0)  cos  (0'  -  0)  w 

the  angles  0  and  0'  being  connected  by  (3). 

There  will  be  equilibrium  if  Q  be  acted  on  by  any  force 
whose  magnitude  lies  between  Px  and  P2. 


FRICTION  OF  A    TRUNNION. 


159 


Fig.48 


98.  Friction  on  Two  Inclined  Planes. — A  beam 
rests  on  two  rough  inclined  planes;  find  the  position  of 
equilibrium. 

Let  a  and  b  be  the  segments,  AG 
and  BG,  of  the  beam ;  let  6  be  the 
inclination  of  the  beam  to  the  hori- 
zon, a  and  ]3  the  inclinations  of  the 
planes,  and  R  and  S  the  total  resist- 
ances. Suppose  that  A  is  on  the 
point  of  ascending  ;  then  the  total 
resistances,  R  and  8,  must  each 
make  the  angle  of  friction  with  the  corresponding  normal 
and  act  to  the  right  of  the  normal. 

The  three  forces,  W,  R,  S,  must  meet  in  a  point  0  (Art. 
62) ;  and  the  angles  GOA  and  GOB  are  equal  to  a  -f-  0, 
and  /J  —  <p,  respectively. 

Hence     (a  +  b)  cot  BGO  =  a  cot  GOA  —  b  cot  GOB, 

or      (a  +  b)  tan  0  =  a  cot  (a  +  <p)  —  b  cot  (0  —  <p).     (1) 

Cok. — If  the  planes  are  smooth,  0  =  0,  and  (1)  becomes 

(a  -f  b)  tan  6  =  a  cot  a  —  b  cot  j3. 
(See  Ex.  7,  Art.  62.) 

99.  Friction  of  a  Trunnion.* — Trunnions  are  the 
cylindrical  projections  from  the  ends  of  a  shaft,  which  rest 
on  the  concave  surfaces  of  cylindrical  boxes.  A  shaft  rests 
in  a  horizontal  position,  with  its  trunnions  on  rough 
cylindrical  surfaces;  find  the  resistance  due  to  friction 
which  is  to  be  overcome  when  the  shaft  begins  to  turn 
about  a  horizontal  axis. 


*  Sometimes  called  "  Journal.' 


1G0 


FRICTION  OF  A    PIVOT. 


Fig.49 


Let  bAcl  and  BAED  be  two  right 
sections  of  the  trunnion  and  its  box; 
the  two  circles  are  tangent  to  each 
other  internally.  If  no  rotation 
takes  place  the  trunnion  presses 
upon  its  lowest  point,  H,  through 
which  the  direction  of  the  resulting 
pressure,  R,  passes ;  if  the  shaft 
begins  to  rotate  in  the  direction  AH,  the  trunnion  ascends 
along  the  inclined  surface,  EAB,  in  consequence  of  the 
friction  on  its  bearing,  until  the  force,  S,  tending  to  move 
it  down  just  balances  the  friction,  F.  Kesolying  R  into  a 
normal  force  Arand  a  tangential  one,  S,  we  have,  since  the 
tangential  component  of  R  in  urging  the  trunnion  down 
the  surface  =  the  friction  which  opposes  it. 

S=  F  =  [iJV;    but     R*  =  S*  +  N*} 

or  R?  =  fiW2  +  IP; 

R 


therefore 
and  the  friction 
F 


N  = 


}iR 


Vi+n2' 


R  tan  </> 


or 


VI  +  ^        Vl  +  tan2  0 
F=  R  sin  0. 


(Art.  95), 


Hence,  to  find  the  friction  upon  a  trunnion,  multiply  the 
resultant  of  the  forces  which  act  upon  it  by  the  sine  of  the 
angle  of  friction. 

100.    Friction  of  a   Pivot. — A   heavy  circular  shaft 
rests  in  a  vertical  position,  with  its  end,  which  is  a  circular 


FRICTIOX  OF  A   PIVOT.  161 

section,  on  a  horizontal  plate;  find  the  resistance  due  to 
friction  which  is  to  be  overcome,  when  the  shaft  begins  to 
revolve  about  a  vertical  axis. 

Let  a  be  the  radius  of  the  circular  section  of  the  shaft; 
let  the  plane  of  (r,  6)  be  the  horizontal  one  of  contact 
between  the  end  of  the  shaft  and  the  plate ;  and  let  the 
centre  of  the  circular  area  of  contact  be  the  pole.  Let 
W  =  the  weight  of  the  shaft,  then  the  vertical  pressure  on 

w 

each  unit  of  surface  is  — 5;  and  therefore,  if  r  dr  dd  is  the 


na 


area-element,  we  have 


W 
the  pressure  on  the  element  =  — -2rdrdd; 


W 
.•.    the  friction  of  the  element  =  u  — -0  r  dr  dd. 

The  friction  is  opposed  to  motion,  and  the  direction  of  its 
action  is  tangent  to  the  circle  described  by  the  element ; 
the  moment  of  the  friction  about  the  vertical  axis  through 
the  centre 

_fiWr2drdd 

therefore  the  moment  of  friction  of  the  whole  circular  end 
«*  ft  Wr*  dr  dd       2fi  Wa 


p2v    pc 


7W*  3 


(1) 


and  consequently  varies  as  the  radius.  Hence  arises  the 
advantage  of  reducing  to  the  smallest  possible  dimensions 
the  area  of  the  base  of  a  vertical  shaft  revolving  with  its 
end  resting  on  a  horizontal  bed. 

From  (1)  we  may  regard  the  whole  friction  due  to  the 
pressure  as  acting  at  a  single  point,  and  at  a  distance  from 
the  centre  of  motion  equal  to  two-thirds  of  the  radius  of 


162  EXAMPLES. 

the  base  of  the  shaft.     This  distance  is  called  the  mean 
lever  of  friction. 

When  the  shaft  is  vertical,  and  rests  upon  its  circular  end 
in  a  cylindrical  socket  the  cylindrical  projection  is  called  a 
Pivot. 

EXAMPLES. 

1.  A  mass  whose  weight  is  750  lbs.  rests  on  a  horizontal 
plane,  and  is  pulled  by  a  force,  P,  whose  direction  makes 
an  angle  of  15°  with  the  horizon  ;  determine  P  and  the 
total  resistance,  R,  the  coefficient  of  friction  being  .62. 

Ans.  P  —  413-3  lbs.;  R  =  756-9  lbs. 

2.  Determine  P  in  the  last  example  if  its  direction  is 
horizontal.  Ans.  P  =  465  lbs. 

3.  Find  the  force  along  the  plane  required  to  draw  a 
weight  of  25  tons  up  a  rough  inclined  plane,  the  coefficient 
of  friction  being  -£%,  and  the  inclination  of  the  plane  being 
such  that  7  tons  acting  along  the  plane  would  support  the 
weight  if  the  plane  were  smooth. 

Ans.  Any  force  greater  than  17  tons. 

4.  Find  the  force  in  the  preceding  example,  supposing 
it  to  act  at  the  most  advantageous  inclination  to  the  plane. 

Ans.  15^  tons. 

5.  A  ladder  inclined  at  an  angle  of  60°  to  the  horizon 
rests  between  a  rough  pavement  and  the  smooth  wall  of  a 
house.  Show  that  if  the  ladder  begin  to  slide  when  a  man 
has  ascended  so  that  his  centre  of  gravity  is  half  way  up, 
then  the  coefficient  of  friction  between  the  foot  of  the 
ladder  and  the  pavement  is  -J-  VS. 

6.  A  body  whose  weight  is  20  lbs.  is  just  sustained  on  a 
rough  inclined  plane  by  a  horizontal  force  of  2  lbs.,  and  a 
force  of  10  lbs.  along  the  plane  ;  the  coefficient  of  friction  is 
| ;  find  the  inclination  of  the  plane.      Ans.  2  tan-1  (f-f ). 


EXAMPLES.  163 

7.  A  heavy  body  is  placed  on  a  rough  plane  whose 
inclination  to  the  horizon  is  sin-1  (f),  and  is  connected  by 
a  string  passing  over  a  smooth  pulley  with  a  body  of  equal 
weight,  which  hangs  freely.  Supposing  that  motion  is  on 
the  point  of  ensuing  up  the  plane,  find  the  inclination  of 
the  string  to  the  plane,  the  coefficient  of  friction  being  £. 

Ans.  0  =  2  tan"1  (£), 

8.  A  heavy  body,  acted  upon  by  a  force  equal  in  magni- 
tude to  its  weight,  is  just  about  to  ascend  a  rough  inclined 
plane  under  the  influence  of  this  force ;  find  the  inclination, 
0,  of  the  force  to  the  inclined  plane. 

Ans.  0  =  -  —  i,  or  2(f)  +  i  —  -,  where  i  =  inclination 

of  the  plane,  and  (f>  =  angle  of  friction.     (0  is  here  sup- 
posed to  be  measured  from  the  upper  side  of  the  inclined 

plane).    If  ^  >  2<p  -f-  i,  0  is  negative  and  the  applied  force 

vill  act  towards  the  under  side. 

9.  In  the  first  solution  of  the  last  example,  what  is  the 
magnitude  of  the  pressure  on  the  plane  ? 

Ans.  Zero.     Explain  this. 

10.  If  the  shaft,  (Art.  100),  is  a  square  prism  of  the 
weight  W,  and  rotates  about  an  axis  in  its  centre,  prove 
that  the  moment  of  the  friction  of  the  square  end  varies  as 
the  side  of  the  square. 

11.  If  the  shaft  is  composed  of  two '  equal  circular 
cylinders  placed  side  by  side,  and  rotates  about  the  line  of 
contact  of  the  two  cylinders,  show  that  the  moment  of  the 
friction  of  the  surface  in  contact  with  the  horizontal  plane 

_  32jua  W 
~~      9tt 

12.  What  is  the  least  coefficient  of  friction  that  will 
allow  of  a  heavy  body's  being  just  kept  from  sliding  down 


164  EXAMPLES. 

an  inclined  plane  of  given  inclination,   the   body    (whose 
weight  is  W)  being  sustained  by  a  given  horizontal  force,  P  ? 

W  tan  i  —  P 


Ans. 


W  +  P  tan  t 


13.  It  is  observed  that  a  body  whose  weight  is  known  to 
be  IF  can  be  just  sustained  on  a  rough  inclined  plane  by  a 
horizontal  force  P9  and  that  it  can  also  be  just  sustained  on 
the  same  plane  by  a  force  Q  np  the  plane;  express  the 
angle  of  friction  in  terms  of  these  known  forces. 

PW 

Ans.  Angle  of  friction  =  cos-1 — 

Q  VP2  +  w2 

14.  It  is  observed  that  a  force,  Q19  acting  up  a  rough 
inclined  plane  will  just  sustain  on  it  a  body  of  weight  W, 
and  that  a  force,  Q2,  acting  up  the  plane  will  just  drag  the 
same  body  up ;  find  the  angle  of  friction. 

Ans.  Angle  of  friction  =  sin-1  - —    2      — ^ 

2VW*-QtQ2 

15.  A  heavy  uniform  rod  rests  with  its  extremities  on 
the  interior  of  a  rough  vertical  circle;  find  the  limiting 
position  of  equilibrium. 

Ans.  If  2«  is  the  angle  subtended  at  the  centre  by  the 
rod,  and  A  the  angle  of  friction,  the  limiting  inclination  of 
the  rod  to  the  horizon  is  given  by  the  equation 

sin  2A 
tan  0  = 


cos  2A  +  cos  2cc 


16.  A  solid  triangular  prism  is  placed,  with  its  axis 
horizontal,  on  a  rough  inclined  plane,  the  inclination  of 
which  is  gradually  increased  ;  determine  the  nature  of  the 
initial  motion  of  the  prism. 

Ans.  If  the  triangle  ABO  is  the  section  perpendicular  to 
the  axis,  and  the  side  AB  is  in  contact  with  the  plane,  A 


EXAMPLES.  165 

being  the  lower  vertex,  the  initial  motion  will  be  one  of 
tumbling  if 


P> 


4A 


the  sides  of  the  triangle  being  a,  h,  c,  and  its  area  A.  If  \i 
is  less  than  this  value,  the  initial  motion  will  be  one  of 
slipping. 

17.  A  frustum  of  a  solid  right  cone  is  placed  with  its 
base  on  a  rough  inclined  plane,  the  inclination  of  which 
is  gradually  increased;  determine  the  nature  of  the  initial 
motion  of  the  body. 

Ans.  If  the  radii  of  the  larger  and  smaller  sections  are  R 
and  r,  and  li  is  the  height  of  the  frustum,  the  initial  motion 
will  be  one  of  tumbling  or  slipping  according  as 

47?      W  +  Rr  +  r2 

f1  ^  ^  ~T~  ' 


h     m  +  2Rr  +  3r2 

18.  An  elliptic  cylinder  rests  in  limiting  equilibrium 
between  a  rough  vertical  and  an  equally  rough  horizontal 
plane,  the  axis  of  the  cylinder  being  horizontal,  ai>d  the 
major  axis  of  the  ellipse  inclined  to  the  horizon  at  ai»  angle 
of  45°.     Find  the  coefficient  of  friction. 


vl  4-  2s2 e4 1 

Aiis.  fi  = » 9  e  being  the  eccentricity 

of  the  ellipse. 


CHAPTER    VI. 

THE    PRINCIPLE    OF    VIRTUAL    VELOCITIES* 

101.  Virtual  Velocity. — If  the  point  of  application  of 
a  force  he  conceived  as  displaced  through  an  indefinitely  small 
space,  the  resolved  part  of  the  displacement  in  the  direction 
of  the  force,  is  called  the  Virtual  Velocity  of  the  force  ;  and 
the  product  of  the  force  into  the  virtual  velocity  has  been 
called  the  virtual  moment]  of  the  force. 

Thus,  let  0  be  the  original,  and  A 
the  new  point  of  application  of  the 
force,  P,  acting  in  the  direction  OP, 
and  let  AN  be  drawn  perpendicular  to 
it.  Then  ON  is  the  virtual  velocity  of 
P,  and  P  •  ON"  is  the  virtual  moment.  OA  is  called  the 
virtual  displacement  of  the  point. 

If  the  projection  of  the  virtual  displacement  on  the  line 
of  the  force  lies  on  the  side  of  0  toward  which  P  acts,  as  in 
the  figure,  the  virtual  velocity  is  considered  positive ;  but 
if  it  lies  on  the  opposite  side,  i.  e.,  on  the  action  line  pro- 
longed through  0,  it  is  negative.  The  forces  are  always 
regarded  as  positive ;  the  sign,  therefore,  of  a  virtual  mo- 
ment will  be  the  same  as  that  of  the  virtual  velocity. 

Cor. — If  6  be  the  angle  between  the  force  and  the  virtual 
displacement,  we  have  for  the  virtual  moment, 

P  •  ON  =  P  .  OA  cos  6  =  P  cos  e  •  OA. 

*  The  principle  of  Virtual  Velocities  was  discovered  by  Galileo,  and  was  very 
fully  developed  by  Bernouilli  and  Lagrange. 

t  Sometimes  called  "  Virtual  Work."  Tbe  name  ll  Virtual  Moment "  was  given 
by  Dubamel. 


VIRTUAL    VELOCITIES.  16? 

Now  P  cos  6  is  the  projection  of  the  force  on  the  direction 
of  the  displacement,  and  is  equal  to  OM,  OP  being  the 
force  and  PM  being  drawn  perpendicular  to  OA.  Hence 
we  may  also  define  the  virtual  moment  of  a  force  as  the 
product  of  the  virtual  displacement  of  its  point  of  applica- 
tion into  the  projection  of  the  force  on  the  direction  of  this 
displacement ;  and  this  definition  for  some  purposes  is 
more  convenient  than  the  former. 

Remark.— A  force  is  said  to  do  work  if  it  moves  the  body  to  which 
it  is  applied ;  and  the  work  done  by  it  is  measured  by  the  product  of 
the  force  into  the  space  through  which  it  moves  the  body.  Generally, 
the  work  done  by  any  force  during  an  infinitely  small  displacement  of 
its  point  of  application  is  the  product  of  the  resolved  part  of  the  force 
in  the  direction  of  the  displacement  into  the  displacement ;  and  this 
is  the  same  as  the  virtual  moment  of  the  force. 

102.    Principle    of   Virtual    Velocities.  —  (1)    The 

virtual  moment  of  a  force  is  equal  to  the  sum  of  the  virtual 
moments  of  its  components. 

Let  OR  represent  a  force,  R,  act- 
ing at  0;  and  let  its  components  be 
P  and  Q,  represented  by  OP  and 
OQ.  Let  OA  be  the  virtual  dis- 
placement of  0,  and  let  its  projec- 
tions on  R,  P,  and  Q,  be  r,  p,  and  fig.si 
q,  respectively.     Then   the   virtual 

moments  of  these  forces  are  R  •  r,  P  •  p,  Q  •  q.  Draw  Rn, 
Pm,  and  Qo,  perpendicular  to  OA.  Then  On,  Om,  and  Oo 
(—  mn),  are  the  projections  of  R,  P,  and  Q,  on  the  direc- 
tion of  the  displacement ;  and  hence  (ArtrlOl,  Cor.)  we  have 

R  .  r  =  OA  •  On ; 
P-p=  0A>0m; 
Q  .  q  =  OA  •  mn. 


168  VIRTUAL    VELOCITIES. 

Hence  P  -p  -f  Q  •  q  =  OA  (Om  -f  mn) 

=  OA  •  On  =  R  •  r. 
(See  Minchin's  Statics,  p.  68.) 

(2)  If  there  are  any  number  of  component  forces  we  maj 
compound  them  in  order,  taking  any  two  of  them  first,  and 
finding  the  virtual  moment  of  their  resultant  as  above,  then 
finding  the  virtual  moment  of  the  resultant  of  these  two 
and  a  third,  likewise  the  virtual  moment  of  the  resultant  of 
the  first  three  and  a  fourth,  and  so  on  to  the  last ;  or  we 
may  use  the  potygon  of  forces  (Art.  33).  The  sum  of  the 
virtual  moments  of  the  forces  is  equal  to  the  virtual  dis- 
placement multiplied  by  the  sum  of  the  projections  on  the 
displacement  of  the  sides  of  the  polygon  which  represent 
the  forces  (Art.  101,  Cor.).  But  the  sum  of  these  projec- 
tions is  equal  to  the  projection  of  the  remaining  side  of  the 
polygon,*  and  this  side  represents  the  resultant,  (Art.  33, 
Cor.  1).  Therefore,  the  sum  of  the  virtual  moments  of  any 
number  of  concurring  forces  is  equal  to  the  virtual  moment 
of  the  resultant. 

(3)  If  the  forces  are  in  equilibrium,  their  resultant  is 
equal  to  zero  ;  hence,  it  follows  that  token  any  number  of 
concurring  forces  are  in  equilibrium,  the  sum  of  their 
virtual  moments  =  0. 

This  principle  is  generally  known  as  the  Principle  of 
Virtual  Velocities,  and  is  of  great  use  in  the  solution  of 
practical  problems  in  Statics. 

*  tfrom  the  nature  of  projections  (Anal.  Geom.,  Art.  168),  it  is  clear  that  in  any 
series  of  points  the  projection  (on  a  given  line)  of  the  line  which  joins  the  first  and 
last,  is  eqnal  to  the  sum  of  the  projections  of  the  lines  which  join  the  points,  two 
and  two.  -Thus,  if  the  sides  of  a  closed  polygon,  taken  in  order,  be  marked  with 
arrows  pointing  from  each  vertex  to  the  next  one;  and  if  their  projections  be 
marked  with  arrows  in  the  same  directions,  then,  lines  measured  from  left  to  right 
being  considered  positive,  and  lines  from  right  to  left  negative,  the  sum  of  the  pro- 
jections of  the  sides  of  a  closed  polygon  on  any  right  line  is  zero. 

8 


VIRTUAL   MOMENTS.  169 

103.  Nature  of  the  Displacement. — It  must  bo  care- 
fully observed  that  the  displacement  of  the  particle  ou 
which  the  forces  act  is  virtual  and  arbitrary.  The  word 
virtual  in  Statics  is  used  to  intimate  that  the  displacements 
are  not  really  made,  but  only  supposed,  i.  e.,  they  are  not 
actual  but  imagined  displacements  ;  but  in  the  motion  of  a 
particle  treated  of  in  Kinetics,  the  displacement  is  often 
taken  to  be  that  which  the  particle  actually  undergoes. 
In  Art.  101,  the  displacement  was  limited  to  an  infinitesi- 
mal. In  some  cases,  however,  a  finite  displacement  may  be 
used,  and  it  may  be  even  more  convenient  to  consider  a 
finite  displacement.  But  in  very  many  cases  any  finite  dis- 
placement is  sufficient  to  alter  the  amount  or  direction  of 
the  forces,  so  as  to  j:>revent  the  principle  of  virtual  velocities 
from  being  applicable.  This  difficulty  can  always  be  avoid- 
ed in  practice  by  assuming  the  displacement  to  be  infinitesi- 
mal ;  and  if  the  virtual  displacement  is  infinitesimal  the 
virtual  velocities  are  all  infinitesimal. 

104.  Equation  of  Virtual  Moments.— Let  Pl9  P2, 

Pz,  etc.,  denote  the  forces,  and  6p1,  6p2,  dp3,  etc.,  the  vir- 
tual velocities  j  then  the  principle  of  virtual  velocities  is 
expressed  (Art.  102)  by  the  equation 

PL  .  6Pl  +  P2  •  6p2  +  P3  .  6p3  +  etc.  =  0; 

or  ZPdp  =  0,  (1) 

which  is  called  the  equation  of  virtual  moments.* 

Sch. — If  the  virtual  displacement  is  at  right  angles  to 
the  direction  of  any  force,  it  is  clear  that  dp,  the  virtual 
velocity,  is  equal  to  zero.  Hence,  when  the  virtual  dis- 
placement  is  at  right  angles  to  the  direction  of  the  force, 

*  Or  virtual  work  (See  Art.  101,  Rem.).  This  equation  has  been  made  by  La- 
grange  the  foundation  of  his  great  work  on  Mechanics,  "  Mecauique  Analytique." 
(Price's  Anal.  Mech.,  Vol.  I,  p.  142.) 


170  SYSTEM  OF  PARTICLES. 

the  virtual  moment  of  the  force  =  0,  and  the  force  will  not 
enter  into  the  equation  of  virtual  moments. 

Such  a  virtual  displacement  is  always  a  convenient  one 
to  choose  when  we  wish  to  get  rid  of  some  unknown  force 
which  acts  upon  a  particle  or  system. 

105.  System  of  Particles  Rigidly  Connected.— (1) 

If  a  particle  in  equilibrium,  under  the  action  of  any  forces, 
be  constrained  to  maintain  a  fixed  distance  from  a  given 
fixed  point,  the  force  due  to  the  constraint  (if  any)  is 
directed  towards  the  fixed  point. 

Let  B  be  the  particle,  and  A  the  fixed  point.  Then  it  is 
clear  that  we  may  substitute  for  the  string  or  rigid  rod 
which  connects  B  with  A,  a  smooth  circular  tube  enclosing 
the  particle,  with  the  centre  of  the  tube  at  A.  Now,  in 
order  that  B  may  be  in  equilibrium  inside  the  tube,  it  is 
necessary  that  the  resultant  of  the  forces  acting  upon  it 
should  be  normal  to  the  tube,  i.  e.,  directed  towards  A. 

(2)  Let  there  be  any  number  of 
particles,  mt,  m8,  m3,  etc.,  each 
acted  on  by  any  forces,  P1?  P3,  P3, 
etc.,  and  connected  with  the  others 
by  inflexible  right  lines  so  that  the 
figure  of  the  system  is  invariable. 
Then  each  particle  is  acted  on  by  all 
the  external  forces  applied  to  it,  and 
by  all  the  internal  forces  proceeding  from  the  internal  con- 
nections of  the  particle  with  the  other  particles  of  the 
system.  Thus  the  particle,  m,  is  acted  on  by  P1?  P2,  etc., 
and  by  the  internal  forces  which  proceed  from  its  connec- 
tion with  m1,  m2,  m3,  etc.,  and  which  act  along  the  lines, 
mrn1,  ?nm2,  etc.,  by  (1)  of  this  Article.  Denote  the  forces 
along  the  lines  mmt,  mm2,  mm3,  etc.,  by  t19  t2,  t3,  etc., 
and   their  virtual  velocities   by  6tlf  6t2,  dts,  etc.      Now 


SYSTEM  OF  PARTICLES.  171 

imagine  that  the  system  is  slightly  displaced  so  as  to 
occupy  a  new  position.  Then  (1)  of  Art.  104  gives  us 
for  m9 

pidlh  +  P*ty%  +  etc.  +  tx6tt  +  t26t2  +  etc.  =  0,   (1) 

for  ml9 

^V?4  +  P6fys  +  etc.  +  tt6tx  +  t26t2  -f  etc.  =  0,   (2) 

proceeding  in  this  way  as  many  equations  may  be  formed  as 
there  are  particles  in  the  system. 

Now  it  is  clear  that  tt6t19  and  t26t29  in  (1)  have  contrary 
signs  from  what  they  have  in  (2).  Thus  if  the  system  is 
moved  to  the  right  in  its  displacement,  tt6t19  and  t26t2  will 
be  positive  in  (1)  and  negative  in  (2)  (Art.  101),  and  hence, 
if  we  add  (1)  and  (2)  together,  these  terms  will  disappear ; 
in  the  same  way,  the  virtual  moment  of  the  internal  force 
along  the  line  connecting  m  with  any  other  particle  disap- 
pears by  addition,  and  the  same  is  true  for  the  internal 
force  between  any  two  particles  of  the  system.  Hence, 
adding  together  all  the  equations,  the  internal  forces 
disappear,  and  the  resulting  equation  for  the  whole 
system  is 

SPdp  =  0,  (1) 

and  the  same  result  is  evidently  true  whatever  be  the  num- 
ber of  particles  forming  the  system.  Hence,  if  any  num- 
ber of  forces  in  a  system  are  in  equilibrium,  the  sum  of 
their  virtual  moments  =  0. 

The  converse  is  evidently  true,  that  if  the  sum  of  the 
virtual  moments  of  the  forces  vanishes  for  every  virtual 
displacement,  the  system  is  in  equilibrium. 

The  following  are  examples  which  are  solved  by  the 
principle  of  ^irtual  velocities. 


172  EXAMPLES. 


EXAMPLES. 

1.  Determine  the  condition  of  equilibrium  of  a  heavj 
body  resting  on  a  smooth  inclined  plane  under  the  action 
of  given  forces. 

Let  W  be  the  weight  of  the  body  *R  p 

sustained   on   the  plane   BO    by  the 
force,  P,  making  an  angle,  0,  with  the 


+  w 


plane.      To   avoid   bringing   the   un-     ^^k>1 
known  reaction,  R,  into  our  equation,  Fig.53 

we  make  the  displacement  of  its  point 
of  application  perpendicular  to  its  line  of  action,  (Art.  104, 
Sch.);  hence  we  conceive  0  as  receiving  a  virtual  dis- 
placement, OA,  at  right  angles  to  R,  the  magnitude  of 
which  in  the  present  case  is  unlimited.  Draw  Km  and. 
An  perpendicular  to  W  and  P  respectively,  Om  and.  On 
are  the  virtual  velocities  of  W  and  P,  (Art.  101) ;  and 
W  •  Om  and  P  •  On  are  their  virtual  moments.  Hence  (1) 
of  Art.  104,  gives 

W  •  Om  —  P  •  On  =  0. 

But  0?n  =  OA  sin  a, 

and  On  =  OA  cos  6 ; 

therefore  W  sin  «  —  P  cos  6  =  0}  (1) 

which  agrees  with  Ex.  3,  Art.  41. 

If  the  force  acts  parallel  to  the  plane,  0  =  0,  and  (1) 
becomes 

P  =  W  sin  « ; 

which  agrees  with  Ex.  1,  Art.  41. 


EXAMPLES. 


173 


2.  Suppose  the  plane  in  Ex.  1  to  he  rough,  and  that  the 
body  is  on  the  point  of  being  dragged  up  the  plane,  find 
the  condition  of  equilibrium. 

The  normal  resistance  will  now  be 
replaced  by  the  total  resistance,  R, 
inclined  to  the  normal  at  an  angle 
=  </>,  the  angle  of  friction  (Art.  95, 
Cor. ).  Let  the  virtual  displacement, 
OA,  take  place  perpendicularly  to 
R,  then  (1)  of  Art.  104,  gives 


W  •  Ora  —  P  •  On  =  0. 


But 

and 
therefore 


Om  =  OA  sin  (a  +  </>), 
On  =  OA  cos  ((f)  —  d); 
W  sin  (a  +  0)  =  P  cos  (<f>  —  6)  ; 
which  agrees  with  (3)  of  Art.  96. 

3.  Determine  the  horizontal  force 
which  will  keep  a  particle  in  a  given 
position  inside  a  circular  tube,  (1) 
when  the  tube  is  smooth  and  (2) 
when  it  is  rough. 

(1)  Let  the  virtual  displacement, 
OA,  be  an  infinitesimal,  ==  ds,  along 
the  tube.    Then  since  ds  is  infinites- 
imal the  virtual  velocity  of  R  =  0.     Then  the  equation  of 
virtual  moments  is 

—  W  •  Om  +  P  •  On  =  0. 

Om  =  ds  •  sin  6, 
On  =  ds  •  cos  6 ; 
W-sin  d  =  P.  cos0: 
P  =  W  tan  6. 


Fig.55 


But 
and 

therefore 
or 


174  EXAMPLES. 

(2)  Suppose  the  force,  P,  just  sustains  the  particle  ;  the 
normal  resistance  must  now  be  replaced  by  the  total  resist- 
ance, making  the  angle,  0,  with  the  normal  at  the  right  of 
it.  Take  the  virtual  displacement,  OA',  at  right  angles  to 
the  total  resistance  (Art.  105,  Sch.),  and  let  it  be  as  before, 
an  infinitesimal  ds.     Then  (1)  of  Art.  104,  gives 

—  W-Om  +  P.Ow'  =  0. 

But  Om  =  ds  •  sin  (0  —  0), 

and  On'  =  ds  •  cos  (0  —  0), 

therefore      W  •  sin  (0  —  0)  =  P  •  cos  (0  —  0); 

or  P  =  W  •  tan  (0  —  0). 

Similarly,  if  the  force,  P,  will  just  drag  the  particle  up 
the  tube  we  obtain 

P  =  W  •  tan  (0  +  0). 

4.  Solve  by  virtual  velocities  Ex.  6,  Art.  62. 

Let  the  displacement  be  made  by  diminishing  the  angle 
«,  which  the  beam  makes  with  the  horizontal  plane,  by  da, 
the  ends  of  the  beam  still  remaining  in  contact  with  the 
horizontal  and  vertical  planes.     Then  the  virtual  velocity  of 

T  =  d  •  2a  cos  a  =  —  2a  sin  a  da  \ 
and  that  of 

W  =  da  sin  a  =  a  cos  a  da, 

and  those  of  the  reactions,  R  and  R';  vanish.     Then  the 
equation  of  virtual  moments  is 


EXAMPLES.  175 

—  T  2a  sin  a  da  -f  W  a  cos  a  da  =  0  J 
•  *.    2T  sin  a  =  W  cos  «. 

5.  Solve  Ex.  8,  Art.  62,  by  virtual  velocities. 

Let  the  displacement  be  made  by  increasing  the  angle 
0  by  dd,  the  point,  A,  remaining  in  contact  with  the  wall; 
the  virtual  displacement  of  B  is  at  right  angles  to  the 
direction  of  the  tension,  T,  and  hence  the  virtual  moment 
of  T  is  zero  ;  the  virtual  velocity  of  W  is 

d  (b  cos  (p  —  a  cos  6)  =  a  sin  6  dd  —  b  sin  (/>  d(p. 

Then  (1)  of  Art.  104,  gives 

W  (a  sin  6  dd  —  b  sin  <f>  d(f>)  =s  0; 

.*.    b  sin  <f>  d(j>  =  a  sin  6  dd. 

But  from  the  geometry  of  the  figure  we  have 

b  sin  (f>  =  2a  sin  6; 

•••    b  cos  <f>  d(\>  =  2a  cos  0  d0; 

.'.    2  tan  0  =  tan  6; 

which,  combined  with  (5)  of  Ex.  8,  Art.  62,  gives  us  the 
values  of  sin  6  and  cos  0 ;  and  these  in  (6)  of  that  Ex. 
give  us  the  value  of  x. 

6.  Solve  Ex.  38,  Art.  65,  by  virtual  velocities. 

Since  the  bar  is  to  rest  in  all  positions  on  the  curve  and  the 
peg,  its  centre  of  gravity  will  neither  rise  nor  fall  when  the 
bar  receives  a  displacement,  hence  its  virtual  velocity  will  =  0 ; 
. ' .    etc. 


176  EXAMPLES. 

7.  In  Ex.  4,  Art.  42,  prove  that  (1)  is  the  equation  of 
virtual  moments. 

8.  Find  the  inclination  of  the  beam  to  the  vertical  in 
Ex.  31,  Art.  G5,  by  virtual  velocities. 

9.  Deduce,  by  virtual  velocities,  (1)  the  formula  for  the 
triangle  of  forces  (see  1  of  Art.  32),  and  (2)  the  formula  for 
the  parallelogram  of  forces  (See  1  of  Art.  30). 


CHAPTER     VI  I. 

MACHINES. 

106.  Functions  of  a  Machine. — A  machine,  Static- 
ally,  is  any  instrument  by  means  of  which  we  may  clianye 
the  direction,  magnitude,  and  point  of  application  of  a 
given  force;  and  Kinetically,  it  is  any  instrument  by  means 
of  which  we  may  change  the  direction  and  velocity  of  a 
given  motion. 

In  applying  the  principle  of  virtual  velocities  to  a  system 
of  connected  bodies,  advantage  is  gained  by  choosing  the 
virtual  displacements  in  certain  directions  (Art.  104,  Sch.). 
AYhen  we  use  this  principle  in  the  discussion  of  machines 
the  displacements  which  we  shall  choose  will  be  those  which 
the  different  parts  of  a  machine  actually  undergo  when  it 
is  employed  in  doing  work,  and  instead  of  equations  of 
virtual  work  we  shall  have  equations  of  actual  work ;  and 
in  future  the  principle  of  virtual  velocities  will  often  be 
referred  to  as  the  Principle  of  Work.  (See  Minchin's 
Statics,  p.  383.) 

Every  machine  is  designed  for  the  purpose  of  overcoming 
certain  forces  which  are  called  resistances  ;  and  the  forces 
which  are  applied  to  the  machines  to  produce  this  effect  are 
called  moving  forces.  When  the  machine  is  in  motion, 
every  moving  force  displaces  its  point  of  application  in  its 
own  direction,  while  the  point  of  application  of  a  resistance 
is  displaced  in  a  direction  opposite  to  that  of  the  resistance. 
Hence,  a  moving  force  is  one  whose  elementary  work  *  is 
positive,  and  a  resistance  is  one  whose  elementary  work  is 
negative.     The  moving  force  is,  for  convenience,  called  the 

*  See  Art.  101,  Rem. 


178  MECHANICAL    ADVANTAGE. 

power;  and  because  the  attraction  of  gravity  is  the  most 
common  form  of  the  force  or  resistance  to  be  overcome  it  is 
usually  called  the  weight. 

The  weight  or  resistance  to  be  overcome  may  be  the  earth's  attrac- 
tion, as  in  raising  a  weight ;  the  molecular  attractions  b  tween  the 
particles  of  a  body  as  in  stamping  or  cutting  a  metal,  or  dividing 
wood  ;  or  friction,  as  in  drawing  a  heavy  body  along  a  rough  road. 
The  p;)wer  may  be  that  of  men,  or  horses,  or  the  steam  engine,  »  ;ic., 
and  may  be  just  sufficient  to  overcome  the  resistance,  or  it  may  be  *>i 
excess  of  what  is  necessary,  or  it  may  be  too  small.  If  just  sufficient, 
the  machine,  if  in  motion,  will  remain  uniformly  so,  or  if  it  be  at  rest 
it  will  be  on  the  point  of  moving,  and  the  power,  weight,  and  frictioa 
will  be  in  equilibrium.  If  the  power  be  in  excess,  the  machine  will 
be  set  in  motion  and  will  continue  in  accelerated  motion.  If  the 
power  be  too  small,  it  will  not  be  able  to  move  the  machine  ;  and  if 
it  be  already  in  motion  it  will  gradually  come  to  rest. 

The  general  problem  with  regard  to  machines  is  to  find 
the  relation  between  the  power  and  the  weight.  Some- 
times it  is  most  convenient  that  this  relation  should  be  one 
of  equality,  i.  e.,  that  the  power  should  equal  the  weight. 
Generally,  however,  it  is  most  convenient  that  the  power 
should  be  very  different  from  the  weight.  Thus,  if  a  man 
has  to  lift  a  weight  of  one  ton  hanging  by  a  rope,  it  is  clear 
that  he  cannot  do  it  unless  the  mechanical  contrivance 
provided  enable  him  to  lift  the  weight  by  exercising  a  pull 
of  very  much  less,  say  one  cwt.  When  the  power  is  much 
smaller  than  the  weight,  as  it  is  in  this  case,  which  is  a 
very  common  one,  the  machine  is  said  to  work  at  a  mechan- 
ical advantage.  When,  as  in  some  other  cases,  it  is  desirable 
that  the  power  should  be  greater  than  the  weight,  there  it 
said  to  be  a  mechanical  disadvantage  of  the  machine. 

107.  Mechanical  Advantage. — (1)  Let  P  and  W  be 

the  power  and  weight,  and  p  and  to  their  virtual  velocities 
respectively ;  and  let  friction  be  omitted.  Then  from  the 
equation  of  virtual  work  (Art.  104),  we  have 

PP-Ww  =  0,    or    £=?, 


MECHANICAL  ADVA  XT  AGE.  179 

which  shows  that  the  smaller  P  is  in  comparison  with  \\\ 
the  smaller  w  will  be  in  comparison  with  p.  But  the 
smaller  P  is  in  comparison  with  W,  the  greater  is  the 
mechanical  advantage.  Hence,  the  greater  the  mechanical 
advantage  is  the  less  will  be  the  virtual  velocity  of  the 
weight  in  comparison  with  that  of  the  power.  Now,  if 
motion  actually  takes  place  the  virtual  velocities  become 
actual  velocities;  and  hence  we  have  the  principle  ivhat  is 
gained  in  power  is  lost  in  velocity. 

(2)  There  are  no  cases  in  which  the  weight  and  power 
are  the  only  forces  to  be  considered.  In  every  movement 
of  a  machine  there  will  always  be  a  certain  amount  of  fric- 
tion ;  and  this  can  never  be  omitted  from  the  equation  of 
virtual  work.  There  are  cases,  however,  as  that  of  a  balance 
on  a  knife-edge,  where  the  friction  is  very  small ;  and  for 
these  the  principle,  what  is  gained  in  power  is  lost  in 
velocity,  is  very  approximately  true.  Where  the  friction  is 
considerable  this  is  no  longer  the  case. 

Let  F  and  /  be  the  resistance  of  friction  and  its  virtual 
velocity,  then  the  equation  for  any  machine  will  take  the 
form 

Pp  —  Ww  —  Ff  =  0, 

which  shows  us  that  although  P  can  be  made  as  small  as  we 
wish  by  taking  p  large  enough,  yet  the  mechanical 
advantage  of  diminishing  P  is  restricted  by  the  fact  that  / 
increases  withjt?;  and  therefore  as  P  diminishes  there  is  a 
corresponding  increase  of  the  work  to  be -done  against  fric- 
tion. Hence  if  friction  be  neglected,  there  is  no  practical 
limit  to  the  ratio  of  P  to  W;  but  if  the  friction  be  con- 
sidered, the  advantage  of  diminishing  P  has  a  limit,  since 
if  Pp  remains  the  same,  Wiv  must  decrease  as  Ff  increases; 
I.  e.,  the  work  done  against  friction  increases  with  the 
complexity  of  the  machine  ;  and  thus  puts  a  practical  limit 
to  the  mechanical  advantage  which  it  is  possible  to  obtain 
by  the  use  of  machines. 


180  SIMPLE  MACHINES. 

108.  Simple  Machines. — The  simple  machines,  some- 
times called  the  Mechanical  Powers,  are  generally  enumer- 
ated as  six  in  number  ;  the  Lever,  the  Wheel  and  Axle,  the 
Inclined  Plane,  the  Pulley,  the  Wedge,  and  the  Screw. 
The  Lever,  the  Inclined  Plane,  and  the  Pulley,  may  be 
considered  as  distinct  in  principle,  while  the  others  are 
combinations  of  them. 

The  efficiency*  of  a  machine  is  the  ratio  of  the  useful 
work  it  yields  to  the  whole  amount  of  work  performed  by 
it.  The  useful  work  is  that  which  is  performed  in  over- 
coming useful  resistances,  while  lost  work  is  that  which  is 
spent  in  overcoming  wasteful  resistances.  Useful  resist- 
ances are  those  which  the  machine  is  specially  designed  to 
overcome,  while  the  overcoming  of  wasteful  resistances  is 
foreign  to  its  purpose.  Friction  and  rigidity  of  cords  are 
wasteful  resistances  while  the  zveight  of  the  body  to  be 
lifted  is  the  useful  resistance. 

Let  W  be  the  work  done  by  the  moving. forces,  Wu  the 
useful  and  Wi  the  lost  work  when  the  machine  is  moving 
uniformly.    Then 

W  =  Wu  +  Wh 
and  if  if  denote  the  efficiency  of  the  machine,  we  have 

M  =  —  • 
W 

In  a  perfect  machine,  where  there  is  no  lost  work,  the 
efficiency  is  unity;  but  in  every  machine  some  of  the  work 
is  lost  in  overcoming  wasteful  resistances,  so  that  the 
efficiency  is  always  less  than  unity ;  and  the  object  of  all 
improvements  in  a  machine  is  to  bring  its  efficiency  as  near 
unity  as  possible. 

The  most  noticeable  of  the  wasteful  resistances  are  fric- 
tion and  rigidity  of  cords ;  and  of  these  we  shall  consider 

*  Sometimes  called  7twdulus. 


EQUILIBRIUM    OF   TEE  LEVER.  181 

only  the  first.     The  student  who  wants  information  on  the 
experimental  laws  of  the  rigidity  of   cords    is   refer] 
Weisl     -.."-  Mechanics,  VoL  I,  p.  363. 

109.  The  Lever. — A  lever  is  a  rigid  bar,  straight  or 
curved,  movable  about  a  fixed  axis,  which  is  called  the 
fulcrum.  The  parts  of  the  lever  into  which  the  fulcrum 
divides  it  are  called  the  arms  of  the  lever.  When  the  arms 
are  in  a  straight  line  it  is  called  a  straight  lever;  in  all 
other  cases  it  is  a  bent  lever. 

L  wen  are  divided,  for  convenience,  into  three  kinds, 
according  to  the  position  of  the  fulcrum.  In  the  first  kind 
the  fulcrum  is  between  the  power  and  the  weight  :  in  the 
second  kind  the  weight  acts  between  the  fulcrum  and  the 
power ;  in  the  third  kind  the  power  acts  between  the  ful- 
crum and  the  weight.  In  the  last  kind  the  power  is  always 
greater  than  the  weight. 

A  pair  of  scissors  furnishes  an  example  of  a  pair  of  levers 
of  the  first  kind;  a  pair  of  nut-craekers  of  the  second  kind; 
and  a  pair  of  shears  of  the  third  kind. 

110.  Conditions  of  Equilibrium 
of  the  Lever. — (1)  Without  Friction. 
Let  AB  be  the  lever  and  C  its  fulcrum; 
and  let  the  two  forces,  P  and  IT.  act  in 
the  plane  of  the  paper  at  the  point ..  A 
and  B,  in  the  directions,  AP  and  BW. 
From  C  draw  CD  and  CE  perpendicular  to  the  directions 
of  P  and  Jr.  Let  a  and  3  denote  the  angles  which  the 
directions  of  the  forces  make  with  the  lever.  Then,  taking 
moments  around  C,  we  have 

P*GD=  ir-CE. 

P  perpendicular  on  direction  of  TT 

IT  —  perpendicular  on  direction  of  P  ' 


182  EQUILIBRIUM   OF  THE  LEVER. 

That  is,  the  condition  of  equilibrium  requires  that  the 
poiver  and  weight  should  be  to  each  other  inversely  as  the 
length  of  their  respective  arms  (Art.  46). 

To  find  the  pressure  on  the  fulcrum,  and  its  direction  ; 
let  the  directions  of  the  pressures,  P  aud  W,  intersect  in 
F;  join  0  aud  F;  then,  since  the  lever  is  in  equilibrium 
by  the  action  of  the  forces,  P  and  W,  and  the  reaction  of 
the  fulcrum,  the  resultant  of  P  and  W  must  be  equal  and 
opposite  to  that  reaction,  and  hence  must  pass  through  C 
and  be  equal  to  the  pressure  on  the  fulcrum.  Denote  this 
resultant  by  R,  the  angle  which  it  makes  with  the  lever  by 
6 ;  and  the  angle  AFB  by  w ;  then  we  have  by  (1)  of  Art.  30 

R2  =  P2  +  W2  +  2PWcos  AFB  ; 

or  R2  =  P2  +  W2  +  2PWco8  w,  (2) 

which  gives  the  pressure,  R,  on  the  fulcrum. 

To  find  its  direction  resolve  P,  W,  and  R  parallel  and 
perpendicular  to  the  lever,  and  we  have 

for  parallel  forces,  P  cos  a  —  W  cos  (3—R  cos  (9  =  0; 

for  perpendicular  forces,  P  sin  a  +  W  sin  (3— R  sin  6  =  0; 

by  transposition  and  division  we  get 

tan  e  -  Psina  +  N^sinJ* 

tan  U  ~  P  cos  a-  WcosP'  (3) 

which  gives  the  direction  of  the  pressure. 

Cor. — When  the  lever  is  bent  or  curved  the  condition  of 
equilibrium  is  the  same. 

Solution  by  the  principle  of  virtual  velocities. 

Suppose  the  lever  to  be  turned  round  C  in  the  direction 
of  P  through  the  angle  dOs  into  the  position  ab;  let  p  and 


EQUILIBRIUM   OF  THE  LEVER.  183 

q  be  the  perpendiculars  CD  and  CE  respectively,  then  the 
virtual  velocity  of  P  will  be  (Art.  101), 


Aa  sin  «  =  AC -eft? -sin  a  = 

Similarly,  the  virtual  velocity  of  W  is  —  qdO. 
Hence,  by  the  equation  of  virtual  work  we  have 

P-p-dd  -  W-q-dO  =  0; 

.-.    P  p  =  W*q.  (4) 

which  is  the  same  as  (1). 

(2)  With  Friction. — In  the  above  we  have  supposed  fric- 
tion to  be  neglected  ;  and  if  the  lever  turns  round  a  sharp 
edge,  like  the  scale  beam  of  a  balance,  the  friction  will  be 
exceedingly  small.  Levers,  however,  usually  consist  of  flat 
bars,  turning  about  rounded  pins  or  studs  which  form  the 
f ulcrums,  and  between  the  lever  and  the  pin  there  will  of 
course  be  friction.  To  find  the  friction  let  r  be  the  radius 
of  the  pin  round  which  the  lever  turns ;  then  the  friction 
on  the  pin,  acting  tangentially  to  the  surface  of  the  pin 
and  opposing  motion,  =  R  sin  6  (Art.  99) ;  and  the  virtual 
velocity  of  the  point  of  application  of  the  friction  =  rdO ; 
and  hence  the  virtual  work  of  the  friction  =  R  sin  (p-rdO. 
Hence  the  equation  of  virtual  work  is 

p.pdO  —  W-qdd  —  R  sin  0  rdO  =  0. 

Substituting  the  value  of  R  from  (2),  and  omitting  dd,  wq 
have 

Pp  -  Wq  =  r  sin  0  V~P*  +  W 2  +  2PW cos  w  ;    (5) 

solving  this  quadratic  for  P  we  have 


184  THE   COMMON  BALANCE. 


_  __  w  pq  +  r2  cos  g)  sin2  0 

^2  —  r2  Sin2  0 


xrr     •     ,  Vp2  H-  2#tf  cos  to  -f  <72  —  r2  sin2  0  sin2  a>    .  . 

±  Wrsm  0— ^- " § 2    .  o  , — r ,  (6) 

jtr  —  r2  sm2  0  v  ' 

which  gives  the  relation  between  the  power  and  the  weight 
when  friction  is  considered,  the  upper  or  lower  sign  of 
r  sin  0  being  taken  according  as  P  or  W  is  about  to  pre« 
ponderate. 

Cor. — If  the  friction  is  so  small  that  it  may  be  omitted, 
r  sin  0  =  0,  and  (6)  becomes 


W      p 


(7) 


111.  The  Common  Balance. — In  machines  generally 
the  object  is  to  produce  motion,  not  rest ;  in  other  words 
to  do  work.  The  statical  investigation  shows  only  the  limit 
of  force  to  be  applied  to  put  the  machine  on  the  point  of 
motion,  or  to  give  it  uniform  motion.  For  any  work  to  be 
done,  the  force  applied  must  exceed  this  limit,  and  the 
greater  the  excess,  the  greater  the  amount  of  work  done. 
There  is,  however,  one  class  of  applications  of  the  lever 
where  the  object  is  not  to  do  work,  but  to  produce  equi- 
librium, and  which  are  therefore  specially  adapted  for  treat- 
ment by  statics.  This  is  the  class  of  measuring  machines, 
where  the  object  is  not  to  overcome  a  particular  resistance, 
but  to  measure  its  amount.  The  testing  machine  is  a  good 
example,  measuring  the  pull  which  a  bar  of  any  material 
will  sustain  before  breaking.  The  common  balance  and 
steelyard  for  weighing,  are  familiar  examples. 

The  common  balance  is  an  instrument  for  weighing ;  it 
is  a  lever  of  the  first  kind,  with  two  equal  arms,  with  a 
scale-pan  suspended  from  each  extremity,  the  fulcrum 
being  vertically  above  the  centre  of  gravity  of  the  beam 
when  the  latter  is  horizontal,  and  therefore  vertically  above 


THE   COMMON  BALANCE.  185 

the  centre  of  gravity  of  the  system  formed  by  the  beam,  the 
scale-pans,  and  the  weights  of  the  scale-pans.  The  sub- 
stance to  be  weighed  is  placed  in  one  scale-pan,  and  weights 
of  known  magnitude  are  placed  in  the  other  till  the  beam 
remains  in  equilibrium  in  a  perfectly  horizontal  position, 
in  which  case  the  weight  of  the  substance  is  indicated  by 
the  weights  which  balance  it.  If  these  weights  differ  by 
ever  so  little  the  horizontality  of  the  beam  will  be  disturbed, 
and  after  oscillating  for  a  short  time,  in  consequence  of 
the  fulcrum  being  placed,  above  the  centre  of  gravity  of  the 
system,  it  wTill  rest  in  a  position  inclined  to  the  horizon  at 
an  angle,  the  extent  of  which  is  a  measure  of  the  sensibility 
of  the  balance. 

The  preceding  explanation  represents  the  balance  in  its  simplest 
form;  in  practice  there  are  many  modifications  and  contrivances 
introduced.  Much  skill  has  been  expended  upon  the  construction  of 
balances,  and  great  delicacy  has  been  obtained.  Thus,  the  beam 
should  be  suspended  by  means  of  a  knife-edge,  i.  e.,  a  projecting 
metallic  edge  transverse  to  its  length,  which  rests  upon  a  plate  of 
agate  or  other  hard  substance.  The  chains  which  support  the  scale- 
pans  should  be  suspended  from  the  extremities  of  the  beam  in  the 
same  manner.  The  point  of  support  of  the  beam  (fulcrum)  should  be 
at  equal  distances  from  the  points  of  suspension  of  the  scales ;  and 
when  the  balance  is  not  loaded  the  beam  should  be  horizontal.  We 
can  ascertain  if  these  conditions  are  satisfied  by  observing  whether 
there  is  still  equilibrium  when  the  substance  is  transferred  to  the 
scale  which  the  weight  originally  occupied  and  the  weight  to  that 
which  (he  substance  originally  occupied. 

The  chief  requisites  of  a  good  balance  are  : 

(1)  When  equal  weights  are  placed  in  the  scale-pans  the 
beam  should  be  perfectly  horizontal. 

(2)  The  balance  should  possess  great  sensibility ;  i.  e.,  if 
two  weights  which  are  very  nearly  equal  be  placed  in  the 
scale-pans,  the  beam  should  vary  sensibly  from  its  horizontal 
position. 


186 


REQUISITES    OF  A    GOOD   BALANCE, 


(3)  When   the   balance   is   disturbed   it   should  readily 
return  to  its  state  of  rest,  or  it  should  have  stability. 

112.  To  Determine  the 
Chief  Requisites  of  a  Good 
Balance. — Let  P  and    W  be 

the  weights  in  the  scale-pans  ; 
Q  the  fulcrum  ;  h  its  distance 
from  the  straight  line,  AB, 
which  joins  the  points  of  at- 
tachment of  the  scale-pans  to 

the  beam ;  (I  the  centre  of  gravity  of  the  beam  ;  and  let 
AB  be  at  right  angles  to  00,  the  line  joining  the  fulcrum 
to  the  centre  of  gravity  of  the  beam.  Let  AO  =  CB  =  a; 
OG  =  k ;  w  =  the  weight  of  the  beam  ;  and  6  =  the 
angle  which  the  beam  makes  with  the  horizon  when  there 
is  equilibrium. 

Now  the  perpendicular  from  0 


on  the  direction  of  P  =  a  cos  6  —  h  sin  0; 
"  W=  a  cosd  +  h  sin  0; 

"  tv  =  k  sin  6  ; 


therefore  taking  moments  round  0  we  have 

P  (a  cos  6— h  sin  6)—  W  (a  cos  d  +  h  sin  &)—wk  sin  6  =  0  ; 


tan  6 


(P—W)a 
(P  +  W )  h  +  wJc 


(1) 


This  equation  determines  the  position  of  equilibrium.  The 
first  requisite — the  horizon tality  of  the  beam  when  P  and 
W  are  equal — is  satisfied  by  making  the  arms  equal. 

The  second  requisite  [(2)  of  Art.  Ill],  requires  that,  for 
a  given  value  of  P  —  W,  the  inclination  of  the  beam  to  the 
horizon  must  be  as  great  as  possible,  and  therefore  the  sen- 
sibility is  greater  the  greater  tan  6  is  for  a  given  value  of 
P  —  W;  and  for  a  given  value  of  tan  6  the  sensibility  is 


REQUISITES    OF  A    GOOD   BALANCE.  187 

greater  the  smaller  the  value  of  P  —  W  is ;  hence  (he  sen- 
sibility may  be  measured  by      '     —r,  which  requires  that 

(/» +*)£  +  •* 

be  as  small  as  possible.  Therefore  a  must  be  large,  and  to, 
h,  and  k  must  be  small ;  i.  e,,  the  arms  must  be  long,  the 
beam  light,  and  the  distances  of  the  fulcrum  from  the 
beam  and  from  the  centre  of  gravity  of  the  beam  must  be 
small. 

The  third  requisite,  its  stability,  is  greater  the  greater 
the  moment  of  the  forces  which  tend  to  restore  the  beam  to 
its  former  position  of  rest  when  it  is  disturbed.  If  P=  W 
this  moment  is 

[(P  +  W)  h  +  wTc\  sin  0, 

which  should  be  made  as  large  as  possible  to  secure  the 
third  requisite. 

This  condition  is,  to  some  extent,  at  variance  with  the 
second  requisite.  They  may  both  be  satisfied,  however,  by 
making  (P  -f  IP)  h  +  wk  large  and  a  large  also  ;  i.  e.,  by 
increasing  the  distances  of  the  fulcrum  from  the  beam  and 
from  the  centre  of  gravity  of  the  beam,  and  by  lengthening 
the  arms.  (See  Todhunter's  Statics,  p.  180,  also  Pratt's 
Mechanics,  p.  78.) 

The  comparative  importance  of  these  qualities  of  sews? - 
bility  and  stability  in  a  balance  will  depend  upon  the  use 
for  which  it  is  intended  ;  for  weighing  heavy  weights, 
stability  is  of  more  importance ;  for  use  in  a  chemical 
laboratory  the  balance  must  possess  great  sensibility ;  and 
instruments  have  been  constructed  which  indicate  a  varia- 
tion of  weight  less  than  a  millionth  part  of  the  whole.  In 
a  balance  of  great  delicacy  the  fulcrum  is  made  as  thin  as 
possible  ;  it  is  generally  a  knife-edge  of  hardened  steel  or 
agate,  resting  on  a  polished  agate  plate,  which  is  supported 
on  a  strong  vertical  pillar  of  brass. 


188  THE  STEELYARD. 

113.  The  Steelyard. — This  is  a  kind  of  balance  in 
which  the  arms  are  unequal  in  length,  the  longer  one  being 
graduated,  along  which  a  poise  may  be  moved  in  order  to 
balance  different  weights  which  are  placed  in  a  scale-pan  on 
the  short-arm.  While  the  moment  of  the  substance 
weighed  is  changed  by  increasing  or  diminishing  its  quan- 
tity, its  arm  remaining  constant,  that  of  the  poise  is 
changed  by  altering  its  arm,  the  weight  of  the  poise 
remaining  the  same. 

114.  To   Graduate   the  Common  Steelyard. — (1) 

When  the  point  of  suspension  is  coincident  with  the  centre 
of  gravity. 

Let  AF  be  the  beam  of  the  steel-  (r) 

yard  suspended  about  an  axis  pass-       a        $" F 

ing  through  its  centre  of  gravity,     W). @?1"  'D'  '  tjfJP  "  "1 

C  ;  on  the  arm,  CF,  place  a  mov-      (8)  p 

able  weight,  P ;  then  if  a  weight,      o  Fig.58 

W,  equal  to  P,  is  suspended  from 
A,  the  beam  will  balance  when  P 
on  the  long  arm  is  at  a  distance 

from  C  equal  to  AC.  If  W  equals  twice  the  weight  of  P, 
the  beam  will  balance  when  the  distance  of  P  from  C  is 
twice  AC;  and  so  on  in  any  proportion.  Hence  if  W  is 
successively  1  lb.,  2  lbs.,  3  lbs.,  etc.,  the  distances  of  the 
notches,  1,  2,  3,  4,  etc.,  where  P  is  placed,  are  as  1,  2,  3, 
etc.,  i.  e.9  the  arm  OF  is  divided  into  equal  divisions,  begin- 
ning at  the  fulcrum,  C,  as  the  zero  point. 

(2)  When  the  point  of  suspension  is  not  coincident  with 
the  centre  of  gravity. 

Let  0  be  the  fulcrum,  W  the  substance  to  be  weighed, 
hanging  at  the  extremity,  A,  and  P  the  movable  weight. 
Suppose  that  when  W  is  removed,  the  weight,  P,  placed  at 
B  will  balance  the  long  arm,  CF,  and  keep  the  steelyard  in 
a  horizontal  position ;  then  the  moment  of  the  instrument 


EXAMPLES.  189 

itself,  about  C,  is  on  the  side,  CF,  and  is  equal  to  P«CB. 
Hence,  if  IT  hangs  from  A,  and  P  from  any  point  E,  then 
for  equilibrium  we  must  have 

P.CE  +  P.BC  =  Ff.AOs 

or  P.BE  =  TF.AC; 


W 
BE=^.AO. 


If  we  make  W  successively  equal  to  P,  2P,  3P,  etc.,  then 
the  values  of  BE  will  be  AC,  2AC,  3A0,  etc.,  and  these 
distances  must  be  measured  off,  commencing  at  B  for  the 
zero  point,  and  the  points  so  determined  marked  1,  2,  3,  4> 
etc.  Such  a  steelyard  cannot  weigh  below  a  certain  limit, 
corresponding  to  the  first  notch,  1. 

To  find  the  length  of  the  divisions  on  the  beam,  divide 
BE,  the  distance  of  the  poise  from  the  zero  point,  by  the 
weight,  W,  which  P  balances  when  at  the  point  E.  The 
steelyard  often  has  two  fulcrums,  one  for  small  and  the 
other  for  large  weights. 

EXAMPLES. 

1.  What  force  must  be  applied  at  one  end  of  a  lever 
12  ins.  long  to  raise  a  weight  of  30  lbs.  hanging  4  ins.  from 
the  fulcrum  which  is  at  the  other  end,  and  what  is  the 
pressure  on  the  fulcrum  ?  Am.  10  lbs.  :  20  lbs. 

2.  A  lever  weighs  3  lbs.,  and  its  weight  acts  at  its  middle 
point ;  the  ratio  of  its  arms  is  1  :  3.  If  a  weight  of  48  lbs. 
be  hung  from  the  end  of  the  shorter  arm,  what  weight 
must  be  suspended  from  the  other  end  to  prevent  motion? 

Ans.  15  lbs. 


190  WHEEL   AND  AXLE. 

3.  The  arms  of  a  bent  lever  are  3  ft.  and  5  ft.  and  inclined 
to  each  other  at  an  angle  6  =  150°.  To  the  short  arm  a 
weight  of  7  lbs.  is  applied  and  to  the  long  arm  a  weight  of 
G  lbs.  is  applied.  .Required  the  inclination  of  each  arm  to 
the  horizon  when  there  is  equilibrium. 

Ans.  The  short  arm  is  inclined  at  an  angle  of  18°  22' 
above  the  horizon,  and  the  long  arm  is  inclined  at  an  angle 
of  48°  22'  beloio  the  horizon. 

115.  The  Wheel  and  Axle.— This 
machine  consists  of  a  wheel,  #,  rigidly 
connected  with  a  horizontal  cylinder, 
b,  movable  round  two  trunnions  (Art. 
99),  one  of  which  is  shown  at  c.  The 
power,  P,  is  applied  at  the  circumfer- 
ence of  the  wheel,  sometimes  by  a  cord 
coiled  round  the  wheel,  sometimes  by 
handspikes   as   in    the   capstan,  or  by 

handles  as  in  the  ivindlass ;  the  weight,  W,  hangs  at  the 
end  of  a  cord  fastened  to  the  axle  and  coiled  round  it. 

116.  Conditions  of  Equilibrium  of  the  Wheel  and 
Axle. — (1)  Let  a  and  b  be  the  radii  of  the  wheel  and  axle 
respectively  ;  P  and  W  the  power  and  weight,  supposed  to 
act  by  strings  at  the  circumference  of  the  wheel  and  axle 
perpendicular  to  the  radii  a  and  b.  Then  either  by  the 
principle  of  virtual  velocities  or  by  the  principle  of  momenta 
we  have 

Pa  =  Wb, 

P  _     radius  of  axle  . 

W       radius  of  wheel  * 

It  is  evident  that,  by  increasing  the  radius  of  the  wheel 
or  by  diminishing  the  radius  of  the  axle,  any  amount  of 
mechanical  advantage  may  be  gained.     It  will  also  be  seen 


DIFFERENTIAL    WHEEL   AND   AXLE. 


191 


that  this  machine  is  only  a  modification  of  the  lever ;  the 
peculiar  advantage  of  the  wheel  and  axle  being  that  an  end- 
less series  of  levers  are  brought  into  play.  In  this  respect, 
then,  it  surpasses  the  common  lever  in  mechanical  advan- 
tage. 

In  the  above  we  have  supposed  friction  to  be  neglected, 
or,  what  amounts  to  the  same  thing,  have  assumed  that  the 
trunnion  is  indefinitely  small.  In  practice,  of  course,  the 
trunnion  has  a  certain  radius,  r,  and  a  certain  coefficient  of 
friction.  Calling  R  the  resultant  of  P  and  W,  and  taking 
into  account  the  friction  on  the  trunnion  we  have  for  the 
relation  between  P  and  W 

Pa  =  Wb  +  r  sin  0  VP*  +  W*  +  2PWcobg),      (2) 

w  being  the  angle  between  the  directions  of  P  and  W 
exactly  as  in  Art.  110. 


(2)  Differential  Wheel  and 
Axle. — By  diminishing  b,  the  radius 
of  the  axle,  the  strength  of  the 
machine  is  diminished ;  to  avoid  this 
disadvantage  a  differential  wheel  and 
axle  is  sometimes  employed.  In  this 
instrument  the  axle  consists  of  two 
cylinders  of  radii  b  and  b' ;  the  rope 
is  wound  round  the  former  in  one 
direction,  and  after  passing  under  a 
movable  pulley  to  which  the  weight 
is  attached,  is  wound  round  the  latter  in  the  opposite  direc- 
tion, so  that  as  the  power,  P,  which  is  applied  as  before, 
tangentially  to  the  wheel  of  radius,  a,  moves  in  its  own 
direction,  the  rope  at  b  winds  up  while  the  rope  at  V 
unwinds. 

For  the  equilibrium  of  the  forces  (whether  at  rest  or  in 
uniform  motion),  the  tensions  of  the  rope  in  bm  and  b'n 


Fig.  60 


192 


TOOTHED    WHEELS. 


are  each  equal  to  \  W.    Hence,  taking  moments  round  thi 
centre  of  the  trunnion,  c,  we  have 


Pa  +  ^Wb'  -iWb  =  0; 
•.    Pa  =  iW  (b  -  V), 


(3) 


hence  by  making  the  difference,  b  —  b\  small,  the  power 
can  be  made  as  small  as  we  please  to  lift  a  given  weight. 
Let  the  wheel  turn  through  the  angle  60  \  the  point  of 
application  of  P  will  describe  a  space  =  a6d,  and  the 
weight  will  be  lifted  through  a  space  =  J  (b  —  b')  68, 
which  latter  will  be  very  small  if  b  —  V  is  very  small. 
Therefore,  since  the  amount  of  work  to  be  done  to  raise  the 
weight  to  any  given  height,  is  constant,  economy  of  power 
is  accomplished  by  a  loss  in  the  time  of  performing  the 
work. 

117.  Toothed  Wheels. — Toothed  or  cogged  wheels  are 
wheels  provided  on  the  circumferences  with  projections 
called  teeth  or  cogs  which  interlock,  as  shown  in  the  figure, 
and  which  are  therefore  capable  of  transmitting  force,  so 
that  if  one  of  the  wheels  be  turned  round  by  any  means, 
the  other  will  be  turned  round  also. 

When  the  teeth  are  on  the  sides  of  the  wheel  instead  of 
the  circumference,  they  are  called  crown  wheels.  When 
the  axes  -of  two  wheels  are 
neither  perpendicular  nor 
parallel  to  each  other,  the 
wheels  take  the  form  of 
frustums  of  cones,  and  are 
called  beveled  wheels.  When 
there  is  a  pair  of  toothed 
wheels  on  each  axle  with  the 
teeth  of  the  large  one  on  one 
axle  fitting  between  the  teeth 


TOOTHED    WHEELS.  193 

of  the  small  one  on  the  next  axle,  the  larger  wheel  of  each 
pair  is  called  the  wheel,  and  the  smaller  is  called  the  pinion. 
By  means  of  a  combination  of  toothed  wheels  of  this  kind 
called  a  train  of  wheels,  motion  may  be  transferred  from 
one  point  to  another  and  work  done,  each  wheel  driving 
the  next  one  in  the  series.  The  discussion  of  this  kind  of 
machinery  possesses  great  geometric  elegance  ;  but  it  would 
be  out  of  place  in  this  work.  We  shall  give  only  a  slight 
sketch  of  the  simplest  case,  that  in  which  the  axes  of  the 
wheels  are  all  parallel.  For  the  investigation  of  the  proper 
forms  of  teeth  in  order  that  the  wheels  when  made  shall 
run  truly  one  upon  another  the  student  is  referred  to  other 
works.* 

118.  To  Find  the  Relation  of  the  Power  and 
Weight  in  Toothed  Wheels. — Let  A  and  B  be  the  fixed 
centres  of  the  toothed  wheels  on  the  circumferences  of 
which  the  teeth  are  arranged ;  QCQ  a  normal  to  the  sur- 
faces of  two  teeth  at  their  point  of  contact,  0.  Suppose  an 
axle  is  fixed  on  the  wheel,  B,  and  the  weight,  W9  suspended 
from  it  at  E  by  a  cord  ;  also,  suppose  the  power,  P,  acts  at 
D  with  an  arm  DA ;  draw  Ka  and  Bh  perpendicular  to 
QCQ.  Let  Q  be  the  mutual  pressure  of  one  tooth  upon 
another  at  C;  this  pressure  will  be  in  the  direction  of  the 
normal  QCQ.  Now  since  the  wheel,  A,  is  in  equilibrium 
about  the  fixed  axis,  A,  under  the  action  of  the  forces, 
P  and  Q,  we  have 

P.AD  =  Q.Aa;  (1) 

and  since  the  wheel,  B,  is  in  equilibrium  about  the  fixed 
axis,  B,  under  the  action  of  the  forces,  Q  and  W,  we  have 

W-BE=  Q-Bb.  (2) 

*  See  Goodeve's  Elements  of  Mechanism :  Rankire's  Applied  Mechanics ;  Mose- 
ley's  Engineering;  Willis's  Principles  of  Mechanism;  Collignon's  Statique;  and 
a  Paper  of  Mr.  Airy's  in  the  Camh  PhiL  'AiWMn  VoL  II,  p.  277. 

9 


194  TRAIN  OF   tf    WHEELS. 

Dividing  (1)  by  (2)  we  have 

P^AD  _  Aat 
PT-BE  ~  Bb' 

moment  of  P Aa 

moment  of  W  ' "  Bb 


or 


If  the  direction  of  the  normal,  QCQ,  at  the  point  of  con- 
tact, C,  changes  as  the  action  passes  from  one  tooth  to  the 
succeeding,  the  relation  of  P  to  W  becomes  variable.  But, 
if  the  teeth  are  of  such  form  that  the  normal  at  their  point 
of  contact  shall  always  be  tangent  to  both  wheels,  the  lines 
Aa  and  Bb  will  become  radii,  and  their  ratio  constant. 
And  since  the  number  of  teeth  in  the  two  wheels  is  propor- 
tional to  their  radii,  we  have 

moment  of  P  _  number  of  teeth  on  the  wheel  P     .  . 
moment  of  W      number  of  teeth  on  the  wheel  W' 

119.  Relation  of  Power  to  Weight  in  a  Train  of  n 

Wheels. — Let  R19  R2,  R3,  etc.,  be  the  radii  of  the  suc- 
cessive wheels  in  such  a  train ;  rlf  r2,  r3,  etc.,  the  radii  of 
the  corresponding  pinions ;  and  let  P,  P19  P29  P3,  .  .  .  W, 
be  the  powers  applied  to  the  circumferences  of  the  successive 
wheels  and  pinions.  Then  the  first  wheel  is  in  equilibrium 
about  its  axis  under  the  action  of  the  forces  P  and  PT, 
since  the  power  applied  to  the  circumference  of  the  second 
wheel  is  equal  to  the  reaction  on  the  first  pinion,  therefore 

P  x  Rx=  Px  x  rv 

Similarly  Px  x  fi2  =  P2  x  r2 ; 

P8  x  i?3  =  P3  x  r8 ; 

etc.       =       etc.; 

/V-l    X    Rn  =    W  X   Tn, 


EXAMPLES.  195 

Multiplying  these  equations  together  and  omitting  common 
factors,  we  have 

_£_  ?*i    X     1*2,    X     7*3    X   .  »  •  •  ,..  v 

lf""^xi?3x^x *  w 

It  will  be  observed,  in  toothed  gearing,  that  the  smaller 
the  radius  of  the  pinion  as  compared  with  the  wheel,  the 
greater  will  be  the  mechanical  advantage.  There  is,  how- 
ever, a  practical  limit  to  the  size  that  can  be  given  to  the 
pinion,  because  the  teeth  must  be  large  enough  for  strength, 
and  must  not  be  too  few  in  number.  Six  is  generally  the 
least  number  admissible  for  the  teeth  of  a  pinion.  Equa- 
tion (1)  shows  that  by  a  train  consisting  of  a  very  few  pairs 
of  wheels  and  pinions  there  is  an  enormous  mechanical 
advantage.  Thus,  if  there  are  three  pairs,  and  the  ratio  of 
each  wheel  to  the  pinion  is  10  to  1,  then  P  is  only  one 
thousandth  part  of  W;  but  on  the  other  hand,  W  will  only 
make  one  turn  where  P  makes  one  thousand.  Such  trains 
of  wheels  are  very  useful  in  machinery  such  as  hand  cranes, 
where  it  is  not  essential  to  obtain  a  quick  motion,  and 
where  the  power  available  is  very  small  in  comparison  to 
the  weight.     (See  Browne's  Mechanics,  p.  109.) 

EXAMPLES. 

1.  What  is  the  diameter  of  a  wheel  if  a  power  of  3  lbs. 
is  just  able  to  move  a  weight  of  12  lbs.  that  hangs  from  the 
axle,  the  radius  of  the  axle  being  2  ins.?        Ans.  16  ins. 

2.  If  a  weight  of  20  lbs.  be  supported  on  a  wheel  and 
axle  by  a  force  of  4  lbs.,  and  the  radius  of  the  axle  is 
|  in.,  find  the  radius  of  the  wheel.  Ans.  3J  ins. 

3.  A  capstan  is  worked  by  a  man  pushing  at  the  end  of 
a  pole.  He  exerts  a  force  of  50  lbs.,  and  walks  10  ft. 
round  for  every  2  ft.  of  rope  pulled  in.  What  is  the 
resistance  overcome  ?  Ans.  250  lbs. 


196  INCLINED  PLANE. 

4.  An  axle  whose  diameter  is  10  ins.,  has  on  it  two 
wheels  the  diameters  of  which  are  2  ft.  and  2J  ft.  respec- 
tively. Find  the  weight  that  would  be  supported  on  the 
axle  by  weights  of  25  lbs.  and  24  lbs.  on  the  smaller  and 
larger  wheels  respectively.  Ans.  132  lbs. 

120.  The  Inclined  Plane — This  has  already  been 
partly  considered  (Art.  96,  etc.).  Let  the  power,  P,  whose 
direction  makes  an  angle,  0,  with  a  rough  inclined  plane, 
be  employed  to  drag  a  weight,  IF,  up  the  plane.  Then  if 
0  is  the  angle  of  friction  and  i  the  inclination  of  the  plane, 
we  have  from  (3)  of  Art.  96, 

sin  (t  +  «.  () 

*  -  n  Cos  (0  -  0)  K  ' 

If  P  acts  along  the  plane,  0  =  0,  and  (1)  becomes 

P=W^SL+J).  (3) 

COS  0  ' 

If  P  acts  horizontally,  0  =  —  %,  and  (1)  becomes 

P  =  FT  tan  {i  +  0).  (3) 

Cor. — If  we  suppose  the  friction  =  0,  (1),  (2),  and  (3) 
become  respectively 

J>=W^,  (4) 

cos  0 

P=  IF  sine,  (5) 

P  =  W  tan  t.  (6) 

Sch. — It  follows  from  (4),  (5),  and  (6)  that  the  smaller 


THE  PULLET.  197 

the  inclination*  of  the  plane  to  the  horizon,  the  greater  will 
be  the  mechanical  advantage.     If  we  take  in  friction  there 

is   an   exception   to    this    rule    when    i '  >  -  —  0.      The 

gradients  on  railways  are  the  most  common  examples  of 
the  use  of  the  inclined  plane ;  these  are  always  made  as  low 
as  is  convenient  in  order  to  enable  the  engine  to  lift  the 
heaviest  possible  train. 

121.  The  Pulley. — The  pulley  consists  of  a  grooved 
wheel,  capable  of  revolving  freely  about  an  axis,  fixed  into 
a  framework,  called  the  block.  A  cord  passes  over  a  por- 
tion of  the  circumference  of  the  wheel  in  the  groove. 
When  the  axis  of  the  pulley  is  fixed,  the  pulley  is  called  a 
fixed  pulley,  and  its  only  effect  is  to  change  the  direction 
of  the  force  exerted  by  the  cord ;  but  where  the  pulley  can 
ascend  and  descend  it  is  called  a  movable  pulley,  and  a 
mechanical  advantage  may  be  gained.  Combinations  of 
pulleys  may  be  made  in  endless  variety;  we  shall  consider 
only  the  simple  movable  pulley  and  three  of  the  more 
ordinary  combinations.  No  account  will  be  here  taken  of 
the  weight  of  the  pulleys  or  of  the  cord,  or  of  friction  and 
stiffness  of  cords.  The  weight  of  a  set  of  pulleys  is  gener- 
ally small  m  comparison  with  the  loads  which  they  lift  ; 
and  the  friction  is  small.  The  use  of  the  pulley  is  to 
diminish  the  effects  of  friction  which  it  does  by  transferring 
the  friction  between  the  cord  and  circumference  of  the 
wheel  to  the  axis  and  its  supports,  which  may  be  highly 
polished  or  lubricated.  The  mechanical  principle  involved 
in  all  calculations  with  respect  to  the  pulley  is  the  constancy 
of  the  force  of  tension  in  all  parts  of  the  same  string 
(Art.  40). 

*  To  find  the  inclination  of  the  plane  for  a  maximum  value  of  P  when  it  acts 
parallel  to  the  plane  we  put  the  derivative  of  P  with  respect  to  i  =  0,  and  get 

-?  =  W  cos  (*_ti>  _  o, .-.  i  =  -  -  <v    Hence  while  the  inclination  of  the  plane 
di  cos  0  '  2 

is  diminishing  from  \  to  \  —  0,  mechanical  advantage  is  diminishing. 
*      2 


198 


FIRST  SYSTEM   OF  PULLEYS. 


122.  The  Simple  Movable  Pulley.— Let  0   be  the 

centre  of  the  pulley  which  is  supported  by  a  cord  passing 
under  it  with  one  end  attached  to  a  beam  at  A  and  the 
other  end  stretched  by  the  force  P. 

Now  since  the  tension  of  the  string, 
ABDP,  is  the  same  throughout,  and  the 
weight,  W,  is  supported  by  the  two 
strings  at  B  and  D,  in  each  of  which 
the  tension  is  P,  we  have 


^ ? 


2P=  W; 


P 

W 


© 


The  same  result  follows  by  the  prin- 
ciple of  virtual  velocities.     Suppose  the  Fig.62 
pulley  -and  the  weight,  W,  to  rise  any 
distance.      Then  it  is  clear  that  both  halves  of  the  string 
must   be   shortened   by   the   same  distance,  and  hence  JP 
must  rise  double  the  distance.;  and  therefore  the  equatiou 
of  virtual  work  gives 

P       1 

The  mechanical  advantage  with  a  single  movable  pulley 

is  2. 


123.  First  System  of  Pulleys,  in  which 
the  same  cord  passes  round  all  the  Pul- 
leys.— In  this  system  there  are  two  blocks,  A 
and  B,  the  upper  of  which  is  fixed  and  the 
lower  movable,  and  each  containing  a  number 
of  pulleys,  each  pulley  being  movable  round 
the  axis  of  the  block  in  which  it  is.  A  single 
corcl  is  attached  to  the  lower  block  and  passes 
alternately  round  the  pulleys  in  the  upper  and 
lower  blocks,  the  portions  of  the  cord  between 
successive  pulleys  being  parallel.    The  portion 


IIIiST  SYSTEM   OF  PULLEYS.  199 

of  cord  proceeding  from  one  pulley  to  the  next  is  called  a 
ply  j  the  portion  at  which  the  power,  P,  is  applied  is 
called  the  tackle-fall. 

Since  the  cord  passes  round  all  the  pulleys  its  tension  is 
the  same  throughout  and  equal  to  P.  Then  if  n  be  the 
number  of  plies  at  the  lower  block,  nP  will  be  the  resultant 
upward  tension  of  the  cords  at  the  lower  block,  which 
must  equal  W ; 

.-.    nP  =  W, 

P       1 

or  Tar  =  — 

W       n 

This  result  follows  also  by  the  principle  of  virtual  veloci- 
ties. Let^?  denote  the  length  of  the  tackle-fall  and  x  the 
common  length  of  the  plies  ;  then  since  the  length  of  the 
cord  is  constant,  we  have 

p  +  nx  =  constant ; 

.*.   dp  -f-  ndx  =  0. 

But  the  equation  of  virtual  work  is 

Pdp  +  Wdx  =  0; 

jy  W  PI 

n  W      n 

This  system  is  most  commonly  used  on  account  of  its 
superior  portability  and  is  the  only  one  of  practical  impor- 
tance. The  several  pulleys  are  usually  mounted  on  a  com- 
mon axis,  as  in  the  figure,  the  cord  being  inclined  slightly 
aside  to  pass  from  one  pair  of  pulleys  to  the  next. 

This  forms  what  is  called  a  set  of  Blocks  and  Falls.  It 
is  very  commonly  used  on  shipboard  and  wherever  weights 
have  to  be  lifted  at  irregular  times  and  places.  The  weight 
of  the  lower  set  of  pulleys  in  this  case  merely  forms  part  of 
the  gross  weight  W. 


200  SECOND   SYSTEM   OF  PULLEYS. 

The  friction  on  the  spindle  of  any  particular  pulley  is 
proportional  to  the  total  pressure  on  the  pulley,  which  is 
clearly  2P.  Hence,  if  \i  is  the  coefficient  of  friction,  the 
resistance  of  friction  on  any  pulley  =  2Pfi ;  and  the 
amount  of  its  displacement,  when  W  is  raised,  will  be  to 
the  displacement  of  W  in  the  ratio  of  the  radius  of  the 
spindle  to  that  of  the  pulley. 

124.  Second  System  of  Pulleys, 
in  which  each  Pulley  hangs  from  a 
fixed  block  by  a  separate  String. — 

Let  A  be  the  fixed  pulley,  n  the  number 

of  movable  pulleys  ;  each  cord  has  one 

end  attached  to  a  fixed  point  in  the  beam, 

and  all  except  the  last  have  the  other  end 

attached  to  a  movable  pulle}',  the  por-  Fig.  64 

tions  not  in  contact  with  any  pulley  being  all  parallel. 

Then  the  tension  of  the  cord  passing  under  the  first 

W 
(lowest)  pulley  =  —  (Art.  122)  ;  the  tension  of  the  cord 

w 

passing  under  the  second  pulley  =  -— ,  and  so  on  ;  and  the 

W 

tension  of  the  cord  passing  under  the  wth  pulley  =  -^, 

which  must  equal  the  power,  P; 


P        1 


(1) 


The  same  result  follows  by  the  principle  of  work.  Sup- 
pose the  first  pulley  and  the  weight  W  to  rise  any  distance, 
x ;  then  it  is  clear  that  both  portions  of  the  cord  passing 
round  this  pulley  will  be  shortened  by  the  same  distance, 
and  hence  the  second  pulley  must  rise  double  this  distance 
or  2x,  and  the  third  pulley  must  rise  double  the  distance  of 
the  second  or  22x,  and  so  on  ;  and  the  nth.  pulley  must  rise 
2n~1x  and  P  must  descend  2nx :  therefore  the  work  of  P 


THIRD   SYSTEM   OF  PULLEYS. 


201 


is  P2nx,  and  the  work  to  be  done  on  W  is  W'X.     Hence 
the  equation  of  work  gives 

P        X 

W~~  2* 


P.  %nx  =    WX, 


I 


r 


125.  Third  System  of  Pulleys,  in  which  each  cord 
is  attached  to  the  -weight. — In  this  system  one  end  of 
each  cord  is  attached  to  the  bar  from  which  the  weight 
hangs,  and  the  other  supports  a  pulley,  the  cords  being  all 
parallel,  and  the  number  of  movable  pulleys  one  less  than 
the  number  of  cords. 

Let  n  be  the  number  of  cords;  then  the 
tension  of  the  cord  to  which  P  is  attached  is 
P ;  the  tension  of  the  second  cord  is  2P  (Art. 
122) ;  that  of  the  next  22P,  and  so  on ;  and 
the  tension  of  the  ^th  cord  is  2n~1P.  Then 
the  sum  of  all  the  tensions  of  the  cords 
attached  to  the  weight  must  equal  W. 
Hence 


P  +  2P  +  2«P  + 


2* 


"5w 

Fig-65 

IP  =  (2»  -  1)  P  =  W; 


P 

W 


2n  —  1 


In  this  system  the  weights  of  the  movable  pulleys  assist  P ; 
in  the  two  former  systems  they  act  against  it. 


EXAMPLES. 

1.  What  force  is  necessary  to  raise  a  weight  of  480  lbs. 
by  an  arrangement  of  six  pulleys  in  which  the  same  string 
passes  round  each  pulley  ?  Ans.  80  lbs. 

2.  Find  the  power  which  will  support  a  weight  of 
800  lbs.  with  three  movable  pulleys,  arranged  as  in  the 
second  system.  Ans.  100  lbs. 


202 


THE    WEDGE. 


3.  If  there  be  equilibrium  between  P  and  W  with  three 
pulleys  in  the  third  system,  what  additional  weight  can  be 
raised  if  2  lbs.  be  added  to  P?  Ans.  14  lbs. 

126.  The  Wedge. — The  wedge  is  a  triangular  prism, 
usually  isosceles,  and  is  used  for  separating  bodies  or  parts 
of  the  same  body  by  introducing  its  edge  between  them  and 
then  thrusting  the  wedge  forward.  This  is  effected  by  the 
blow  of  a  hammer  or  other  such  means,  which  produces  a 
violent  pressure,  for  a  short  time,  in  a  direction  perpen- 
dicular to  the  back  of  the  wedge,  and  the  resistance  to  be 
overcome  consists  of  friction  and  a  reaction  due  to  the 
molecular  attractions  of  the  particles  of  the  body  which 
are  being  separated.  This  reaction  will  be  in  a  direction 
perpendicular  to  the  inclined  surface  of  the  wedge. 

127.  The    Mechanical  Ad- 
vantage of  the  Wedge. — Let 

ACB  represent  a  section  of  the 
wedge  perpendicular  to  its  in- 
clined faces,  the  wedge  having 
been  driven  into  the  material  a 
distance  equal  to  DO  by  a  force, 
P,  acting  in  the  direction  DO. 
Draw  DE,  DF,  perpendicular  to 
AC,  BO,  and  let   R  denote  the 

reactions  along  ED  and  FD  ;  then  \iR  will  be  the  friction 
acting  at  E  and  F  in  the  directions  EA  and  FB.  Let  the 
angle  of  the  wedge  or  ACB  =  2a. 

Resolve  the  forces  which  act  on  the  wedge  in  directions 
perpendicular  and  parallel  to  the  back  of  the  wedge,  then 
we  have  for  perpendicular  forces 


Fig.66 


P  =  2R  sin  a  -f  2fiR  cos  es. 


(i) 


TJiis  equation  may  also  he  obtained  from  the  principle  of 
work  as  follows :    If  the  weJge  has  been  driven  into  the 


MECHANICAL  ADVANTAGE   OF    WEDGE.  203 

material  a  distance  equal  to  DC  by  a  force,  P,  acting  in 
the  direction  DO,  then  the  work  done  by  P  is  P  x  DO 
(Art.  101,  Rem.);  and  since  the -points  E  and  F  were 
originally  together,  the  work  done  against  the  resistance 
R  is  R  x  DE  +  R  x  DF  =  2R  x  DE ;  and  the  work 
done  against  friction  is  2\iR  x  EC.  Hence  the  equation 
of  woik  is 

P  x  DC  =  %R  x  DE  +  2fiR  x  EC,  (2) 

which  reduces  to  (1)  by  substituting  sin  a  and  cos  a  for 
DE  EC 

DC  and  DC" 

Cor. — If  friction  be  neglected,  (2)  becomes 

P  _  2DE  _  A  B 
R  ~  DC    ~  AC 

,      .  P back  of  the  wedge 

R  ~  length  of  one  of  the  equal  sidles 

It  follows  that  the  narrower  the  back  of  the  wedge,  the 
greater  will  be  the  mechanical  advantage.  Knives,  chisels, 
and  many  other  implements  are  examples  of  the  wedge. 

In  the  action  of  the  wedge  a  great  part  of  the  power  is 
employed  in  cleaving  the  material  into  which  it  is  driven. 
The  force  required  to  effect  this  is  so  great  that  instead  of 
applying  a  continuous  pushing  force  perpendicular  to  the 
back  of  the  wedge,  it  is  driven  by  a  series  of  blows.  Be- 
tween the  blows  there  is  a  powerful  reaction,  R,  acting  to 
push  the  wedge  back  again  out  of  the  cleft,  and  this  is 
resisted  by  the  triction  which  now  acts  in  the  directions 
EC  and  FC.  Hence  when  the  wedge  is  on  the  point  of 
starting  back,  between  the  blows,  the  equation  of  equi- 
librium will  be  from  (1) 

2R  sin  a  —  %\iR  cos  a  =  0 ; 

.  • .     a  =  tan-1  fi. 


204 


THE  SCREW. 


And  the  wedge  will  fly  back  or  not  according  as  a  >  01 
<  tan-1  p.  (See  Browne's  Mechanics,  p.  117.  Also  Magnus's 
Mechanics,  p.  157.) 

128.  The  Screw. — The  screw  consists  of  a  right  -cir- 
cular cjdinder,  on  the  convex  surface  of  which  there  is 
traced  a  uniform  projecting  thread,  abed  ....  inclined  at 
a  constant  angle  to  straight  lines  parallel  to  the  axis  of  the 
cylinder.  The  path  of  the  thread 
may  be  traced  by  the  edge  AC  of 
an  inclined  plane,  ABC,  wrapped 
round  the  c}7linder;  the  base  of 
the  plane  corresponding  with  the 
circumference  of  the  cylinder,  and 
the  height  of  the  plane  with  the 
distance  between  the  threads  which 
is  called  the  pitch  of  the  screw. 
The  threads  may  be  rectangular  or 
triangular  in  section.  The  cylinder 
fits  into  a  block,  on  the  inner  sur- 
face of  which  is  cut  a  groove  which  is  the  exact  counterpart 
of  the  thread.  The  block  in  which  the  groove  is  cut  is  often 
called  the  nut  The  power  is  generally  applied  at  the  end  of 
a  lever  fixed  to  the  centre  of  the  cylinder,  or  fixed  to  the  nut. 
It  is  evident  that  a  screw  never  requires  any  pressure  in  the 
direction  of  its  axis,  but  must  be  made  to  revolve  only ; 
and  this  can  be  done  by  a  force  acting  at  right  angles  to 
the  extremities  of  its  diameter,  or  its  diameter  produced. 

129.  The  Relation  between  the  Power  and  the 
"Weight  in  the  Screw.— Suppose  the  power,  P.  to  act  in 
a  plane  perpendicular  to  the  axis  of  the  cylinder  and  at  the 
end  of  an  arm,  DE  =  a,  and  suppose  the  screw  to  have 
made  one  revolution,  the  power,  P,  will  have  moved 
through  the  circumference  of  which  a,  is  the  radius,  and 
the  work  done  by  P  will  be  Px2na.     During  the  same 


Fig.67 


THE  SCREW.  205 

time  the  screw  will  have  moved  in  the  direction  of  its  axis 
through  the  distance,  AB  =  2nr  tan  «,  r  being  the  radius 
of  the  cylinder,  and  «  the  angle  which  the  thread  of  the 
screw  makes  with  its  base.  Then  as  this  is  the  direction  in 
which  the  resistance  is  encountered,  the  work  done  against 
the  resistance,  W,  is  W'Zttt  tan  «.  Hence  if  no  work  is  lost 
the  equation  of  work  will  be 

P  x  2na  =  W  x  2nr  tan  a.  (1) 

That  is  the  power  is  to  ihe  weight  as  the  pitch  of  the  screw 
is  to  the  circumference  described  by  the  power. 

If  there  is  friction  between  the  thread  and  the  groove,  let 
R  be  the  normal  pressure  at  any  point,  p,  of  the  thread, 
and  fiR  the  friction  at  this  point,  then  the  work  done 
against  the  friction  in  one  revolution  is  ^R  2  tit  sec  «,  2R 
denoting  the  sum  of  the  normal  reactions  at  all  points  of 
the  thread.     Hence  the  equation  of  work  is 

P  2na  =  W2nr  tan  a  -f  fi  2nr  sec  a^R.  (2) 

But,  for  the  equilibrium  of  the  screw,  resolving  parallel 
to  the  axis,  we  have 

W  =  2  (R  cos  a  —  pR  sin  a), 

W 

therefoie  ZR  = -. ; 

cos  a  —  fi  sin  a 

which  in  (2)  gives 

n         _  ,  fir  sec  a  W 

Pa  =  Wr  tan  a  +  — — = ; 

cos  a  —  \i  sin  a 

or  Pa  =  Wr  tan  («  +  0),  (3) 

(t>  being  the  angle  of  friction. 


206 


prony's  differential  screw. 


129a.  Prony's  Differential  Screw. — If  h  denote  the 
pitch  of  a  screw  (1)  becomes 


2Prra 


Wh, 


which  expresses  the  relation  between  P  and  W,  when  fric- 
tion is  neglected.  Therefore  the  mechanical  advantage  is 
gained  by  making  the  pitch  very  small.  In  some  cases, 
however,  it  is  desirable  that  the  screw  should  work  at  fair 
speed,  as  in  ordinary  bolts  and  nuts,  and  then  the  pitch 
must  not  be  too  small.  In  cases  where  the  screw  is  used 
specially  to  obtain  pressure,  as  in  screw-presses  for  cotton, 
etc.,  we  do  not  care  for  speed,  but  only  for  pressure.  But 
in  practice  it  is  impossible  to  get  the  pitch  very  small  from 
the  fact  that  if  the  angle  of  inclination  is  very  flat,  the 
threads  run  so  near  each  other  as  to  be  too  weak,  in. which 
case  the  screw  is  apt  to  "  strip  its  thread/'  that  is,  to  tear 
bodily  out  of  the  hole,  leaving  the  thread  behind. 

Where  very  great  pressure  is  required  a  differential  nut- 
Ziole  is  resorted  to.  Let  the  screw  work  in  two  blocks, 
A  and  B,  the  first  of 
which  is  fixed  and  the 
second  movable  along  a  PC 

fixed  groove,  n ;  and  let 
h  be  the  pitch  of  the     — ■= 
thread  which  works  in  Fig. 68 

the  block,  A,  and  h'  the  pitch  of  the  thread  which  works 
in  the  block  B.  Then  one  revolution  of  the  screw  impresses 
two  opposite  motions  on  the  block,  B,  one  equal  to  h  in  the 
direction  in  which  the  screw  advances,  and  the  other  equal 
to  h'  in  the  opposite  direction.  If  then  the  block,  B,  is 
connected  with  the  resistance  W,  we  have  by  the  principle 
of  work 

2Pna  =  W(h-h')\ 

and  the  requisite  power  will  be  diminished  by  diminishing 


EXAMPLES.  20? 

//  —  h'.  By  means  of  this  screw  a  comparatively  small 
pressure  may  be  made  to  yield  a  pressure  enormously 
greater  in  magnitude. 

examples. 

1.  A  lever  10  ins.  long,  the  weight  of  which  is  4  lbs.,  and 
acts  at  its  middle  point,  balances  about  a  certain  point 
when  a  weight  of  6  lbs.  is  hung  from  one  end;  find  the 
point.  Ans.  2  ins.  from  the  end  where  the  weight  is. 


2.  A  lever  weighing  8  lbs.  balances  at  a  point  3  ins.  from 
one  end  and  9  ins.  from  the  other.  Will  it  continue  to  bal- 
ance about  that  point  if  equal  weights  be  suspended  from 
the  extremities? 

3.  A  beam  whose  length  is  12  ft.  balances  at  a  point  2  ft. 
from  one  end  ;  but  if  a  weight  of  100  lbs.  be  hung  from  the 
other  end  it  balances  at  a  point  2  ft.  from  that  end ;  find  the 
weight  of  the  beam.  Ans.  25  lbs. 

4.  A  lever  7  feet  long  is  supported  in  a  horizontal  posi- 
tion by  props  placed  at  its  extremities  :  find  where  a  weight 
of  28  lbs.  must  be  placed  so  that  the  pressure  on  one  of  the 
props  may  be  8  lbs.  Ans.  Two  feet  from  the  end. 

5.  Two  weights  of  12  lbs.  and  8  lbs.  respectively  at  the 
ends  of  a  horizontal  lever  10  feet  long  balance :  find  how 
far  the  fulcrum  ought  to  be  moved  for  the  weights  to  bal- 
ance when  each  is  increased  by  2  lbs.      Ans.  Two  inches. 

6.  A  lever  is  in  equilibrium  under  the  action  of  the  forces 
P  and  Q,  and  is  also  in  equilibrium  when  P  is  trebled  and 
Q  is  increased  by  6  lbs.:  find  the  magnitude  of  Q. 

Ans.  3  lbs. 

7.  In  a  lever  of  the  first  kind,  let  the  power  be  217  lbs., 
the  weight  725  lbs.,  and  the  angle  between  them  126°. 
Find  the  pressure  on  the  fulcrum.  Ans.  622.7  lbs. 


208  EXAMPLES. 

8.  If  the  power  and  weight  in  a  straight  lever  of  the 
first  kind  be  17  lbs.  and  32  lbs.,  and  make  with  each  other 
an  angle  of  79° ;  find  the  pressure  on  the  fulcrum. 

Ans.  39  lbs. 

9.  The  length  of  the  beam  of  a  false  balance  is  3  ft. 
9  ins.  A  body  placed  in  one  scale  balances  a  weight  of 
9  lbs.  in  the  other ;  but  when  placed  in  the  other  scale  it 
balances  4  lbs.;  required  the  true  weight,  W,  of  the  body 
and  the  lengths,  a  and  b,  of  the  arms. 

Ans.    W  =  G  lbs.;  a  =  1  ft.  G  ins.;  I  =  2  f t.  3  ins. 

10.  If  a  balance  be  false,  having  its  arms  in  the  ratio  of 
15  to  16,  find  how  much  per  lb.  a  customer  really  pays 
for  tea  which  is  sold  to  him  from  the  longer  arm  at  3s.  9d. 
per  lb.  Ans.  4s.  per  lb. 

11.  A  straight  uniform  lever  whose  weight  is  50  lbs.  and 
length  6  feet,  rests  in  equilibrium  on  a  fulcrum  when  a 
weight  of  10  lbs.  is  suspended  from  one  extremity:  find  the 
position  of  the  fulcrum  and  the  pressure  on  it*. 

Ans.  2J  ft.  from  the  end  at  which  10  lbs.  is  suspended ; 
60  lbs. 

12.  On  one  arm  of  a  false  balance  a  body  weighs  11  lbs.; 
on  the  other  17  lbs.  3  oz.;  what  is  the  true  weight? 

Ans.  13  lbs.  12  oz. 

13.  A  bent  lever  is  composed  of  two  straight  uniform 
rods  of  the  same  length,  inclined  to  each  other  at  120°,  and 
the  fulcrum  is  at  the  point  of  intersection :  if  the  weight  of 
one  rod  be  double  that  of  the  other,  show  that  the  lever  will 
remain  at  rest  with  the  lighter  arm  horizontal. 

14.  A  uniform  lever,  I  feet  long,  has  a  weight  of  W  lbs., 
suspended  from  its  extremity;  find  the  position  of  the  ful- 
crum when  the  long  end  of  the  lever  balances  the  short 


EXAMPLES.  209 

end  with  the  weight  attached  to  it,  supposing  each  unit  of 
length  of  the  lever  to  be  w  lbs. 

Ans.   rt  ,  ...  .    7   .  is  the  short  arm. 
2  ( It  -f  ho) 

15.  A  lever,  I  ft.  long,  is  balanced  when  it  is  placed  upon 
a  prop  I  of  its  length  from  the  thick  end;  when  a  weight 
of  W  lbs.  is  suspended  from  the  small  end  the  prop  must 
be  shifted  £  ft.  towards  it  in  order  to  maintain  equilibrium; 
required  the  weight  of  the  lever.  Arts.  \  W. 

16.  A  lever,  I  ft.  long;  is  balanced  on  a  prop  by  a  weight 
of  W  lbs.;  first,  when  the  weight  is  suspended  from  the 
thick  end  the  prop  is  a  ft.  from  it;  secondly,  when  the 
weight  is  suspended  from  the  small  end  the  prop  is  b  ft. 
from  it ;  required  the  weight  of  the  lever. 

^^L±Alb, 

1  —  (a  +b) 

17.  The  forces,  P  and  W,  act  at  the  arms,  a  and  b, 
respectively,  of  a  straight  lever.  When  P  and  W  make 
angles  of  30°  and  90°  with  the  lever,  show  that  when  equi- 
librium takes  place  P  =  — — 

18.  Supposing  the  beam  of  a  false  balance  to  be  uniform, 
a  and  b  the  lengths  of  the  arms,  P  and  Q  the  apparent 
weights,  and  W  the  true  weight;  when  the  weight  of  the 
beam  is  taken  into  account  show  that 

a       P  —W 


b        W-Q 

19.  If  a  be  the  length  of  the  short  arm  in  Ex.  14,  what 

must  be  the  length  of  the  whole  lever  when  equilibrium 

takes  place ?  /2aW 

Ans.  a  +  \  / f-  a\ 

V    w 

20.  A  man  whose  weight  is  140  lbs.  is  just  able  to  sup- 
j3ort  a  weight  that  hangs  over  an  axle  of  6  ins.  radius,  by 


210  EXAMPLES. 

hanging  to  the  rope  that  passes  over  the  corresponding 
wheel,  the  diameter  of  which  is  4  ft.;  find  the  weight  sup- 
ported. Ans.  560  lbs. 

21.  If  the  difference  between  the  diameter  of  a  wheel  and 
the  diameter  of  the  axle  be  six  times  the  radius  of  the  axle, 
find  the  greatest  weight  that  can  be  sustained  by  a  force  of 
60  lbs.  Ans.  240  lbs. 

22.  If  the  radius  of  the  wheel  is  three  times  that  of  the 
axle,  and  the  string  round  the  wheel  can  support  a  weight 
of  40  lbs.  only,  find  the  greatest  weight  that  can  be  lifted. 

Ans.  120  lbs. 

23.  What  force  will  be  required  to  work  the  handle  of  a 
windlass,  the  resistance  to  be  overcome  being  1156  lbs.,  the 
radius  of  the  axle  being  six  ins.,  and  of  the  handle  2  ft. 
8  ins.?  Ans.  216.75  lbs. 

24.  Sixteen  sailors,  exerting  each  a  force  of  29  lbs.,  push 
a  capstan  with  a  length  of  lever  equal  to  8  ft.,  the  radius  of 
the  capstan  being  1  ft.  2  ins.  Find  the  resistance  which 
this  force  is  capable  of  sustaining. 

Ans.   1  ton  8  cwt.  1  qr.  17f  lbs. 

25.  Supposing  them  to  have  wound  the  rope  round  the 
capstan,  so  that  it  doubles  back  on  itself,  the  radius  of  the 
axle  is  thus  increased  by  the  thickness  of  the  rope.  If  this 
be  2  ins.  how  much  will  the  power  of  the  instrument  be 
diminished.  Ans.  By  \,  or  12 \  per  cent. 

26.  The  radius  of  the  axle  of  a  capstan  is  2  feet,  and  six 
men  push  each  with  a  force  of  one  cwt.  on  spokes  5  feet 
long ;  find  the  tension  they  will  be  able  to  produce  in  the 
rope  which  leaves  the  axle.  Ans.  15  cwt. 

27.  The  difference  of  the  diameters  of  a  wheel  and  axle 
is  2  feet  6  inches ;  and  the  weight  is  equal  to  six  times  the 
power  ;  find  the  radii  of  the  wheel  and  the  axle. 

Ans.  18  ins.;  3  ins. 


EXAMPLES.  211 

28.  If  the  radius  of  a  wheel  is  4  ft..,  and  of  the  axlfe 
8  ins.,  find  the  power  that  will  balance  a  weight  of 
500  lbs.,  the  thickness  of  the  rope  coiled  round  the  axle 
being  one  inch,  the  power  acting  without  a  rope. 

Ans.  88.54  lbs. 

29.  Two  given  weights,  P  and  Q,  hang  vertically  from 
two  points  in  the  rfm  of  a  wdieel  turning  on  an  axis; 
find  the  position  of  the  weights  when  equilibrium  takes 
place,  supposing  the  angle  between  the  radii  drawn  to 
the  points  of  suspension  to  be  90°,  and  that  0  is  the 
angle  which  the  radius,  drawn  to  P's  point  of  sus- 
pension, makes  with  the  vertical.  ,        ,       _        Q 

v  '  Ans.  tan  6  =  ^. 

30.  What  weight  can  be  supported  on  a  plane  by  a  hori- 
zontal force  of  10  lbs.,  if  the  ratio  of  the  height  to  the  base 
is  J?  Ans.  134  lbs. 

31.  The  inclination  of  a  plane  is  30°,  and  a  weight  of 
10  lbs.  is  supported  on  it  by  a  string,  bearing  a  weight  at 
its  extremity,  which  passes  over  a  smooth  pulley  at  its 
summit ;  find  the  tension  in  the  string.  Ans. .  5  lbs. 

32.  The  angle  of  a  plane  is  45° ;  what  weight  can  be 
supported  on  it  by  a  horizontal  force  of  3  lbs.,  and  a  force 
of  4  lbs.  parallel  to  the  plane,  both  acting  together. 

Ans.  3  +  4  V2  lbs. 

33.  A  body  is  supported  on  a  plane  by  a  force  parallel 
to  it  and  equal  to  \  of  the  weight  of  the  body ;  find  the 
ratio  of  the  height  to  the  base  of  the  plane. 

Ans.  1  :  2  a/G. 

34.  One  of  the  longest  inclined  planes  in  the  world  is 
the  road  from  Lima  to  Callao,  in  S.  America ;  it  is  G  miles 
long,  and  the  fall  is  511  ft.     Calculate  the  inclination. 

Ans.  55'  27",  or  1  yard  in  62. 


212  EXAMPLES. 

35.  If  the  force  required  to  draw  a  wagon  on  a  horizontal 
road  be  g^th  part  of  the  weight  of  the  wagon,  what  will  be 
the  force  required  to  draw  it  up  a  hill,  the  slope  of  which 
is  1  in  43.  Ans.  T4.Vitn  Part  °f  the  weight. 

36.  If  the  force  required  to  draw  a  train  of  cars  on  a 
level  railroad  be  %\^\\  part  of  the  load,  find  the  force 
required  to  draw  it  up  a  grade  of  1  irf  56. 

Ans.  4-3-Vsth  part  of  the  load. 

37.  What  force  is  required  (neglecting  friction)  to  roll  a 
cask  weighing  964  lbs.  into  a  cart  3  ft.  high,  by  means  of  a 
plank  14  ft.  long  resting  against  the  cart. 

Ans.  The  force  must  exceed  206^-  lbs. 

38.  A  body  is  at  rest  on  a  smooth  inclined  plane  when 
the  power,  weight  and  normal  pressure  are  18,  26,  and 
12  lbs.  respectively;  find  the  inclination,  a,  of  the  plane  to 
the  horizon,  and  the  angle,  0,  which  the  direction  of  the 
power  makes  with  the  plane. 

Ans.  a  =  37°  21'  26";  0  =  28°  46'  54". 

39.  If  the  power  which  will  support  a  weight  when  act- 
ing along  the  plane  be  half  that  which  will  do  so  acting 
horizontally,  find  the  inclination  of  the  plane.  Ans.  60°. 

40.  A  power  P  acting  along  a  plane  can  support  W,  and 
acting  horizontally  can  support  x ;  show  that 

pa  _  W  2  _  x\ 

41.  A  weight  W  would  be  supported  by  a  power  P  act- 
ing horizontally,  or  by  a  power  Q  acting  parallel  to  the 
plane ;  show  that 

±  =  ±  +  ±. 

Q2  —  p2  -T    W2 

42.  The  base  of  an  inclined  plane  is  8  ft.,  the  height 
6  ft,  and  W  =  10  tons ;  required  P  and  the  normal 
pressure,  N,  on  the  plane. 

Ans.   P  =  6  tons;  N  =  8  tons. 


EXAMPLES.  213 

43.  A  weight  is  supported  on  an  inclined  plane  by  a 
force  whose  direction  is  inclined  to  the  plane  at  an  angle 
of  30°  ;  when  the  inclination  of  the  plane  to  the  horizon  is 
30°,  show  that  W  =  P  V'S. 

44.  A  man  "weighing  150  lbs.  raises  a  weight  of  4  cwt.  by 
a  system  of  four  movable  pulleys  arranged  according  to  the 
second  system  ;  what  is  his  pressure  on  the  ground  ? 

Ans.  122  lbs. 

45.  What  power  will  be  required  in  the  second  system 
kvith  four  movable  pulleys  to  sustain  a  weight  of  17  tons 
12  cwt.  Ans.  1  ton  2  cwt. 

46.  Two  weights  hang  over  a  pulley  fixed  to  the  summit 
of  a  smooth  inclined  plane,  on  which  one  weight  is  sup- 
ported, and  for  every  3  ins.  that  one  descends  the  other 
rises  2  ins.;  find  the  ratio  of  the  weights,  and  the  length 
of  the  plane,  the  height  being  18  ins.    Ans.  2  :  3  ;  27  ins. 

47.  If  W  =  336  lbs.  and  P  =  42  lbs.  in  a  combination 
of  pulleys  arranged  according  to  the  first  system,  how  many 
movable  pulleys  are  there  ?  Ans.  4. 

48.  In  a  system  of  pulleys  of  the  third  kind  in  which 
there  are  4  cords  attached  to  the  weight,  determine  the 
weight,  W,  supported,  and  the  strain  on  the  fixed  pulley, 
the  power  being  100  lbs.,  and  the  weight,  w,  of  each 
pulley  5  lbs. 

Ans.  W  =  15P  +  llzv  =  1555  lbs.;  Strain  =  16P  +  15w 
=  1675  lbs. 

49.  In  a  system  of  pulleys  of  the  third  kind,  there  are 
2  movable  pulleys,  each  weighing  2-J-  lbs.  What  power  is 
required  to  support  a  weight  of  6  cwt.  ?    Ans.  94.57  lbs. 

50.  Find  the  power  that  will  support  a  weight  of  100  lbs. 
by  means  of  a  system  of  4  pulleys,  the  strings  being  all 
attached  to  the  weight,  and  each  pulley  weighing  1  lb. 

Ans.  5|-f  lbs. 


214  EXAMPLES, 

51.  The  circumference  of  the  circle  corresponding  to  the 
point  of  application  of  P  is  6  feet ;  find  how  many  turns 
the  screw  must  make  on  a  cylinder  2  feet  long,  in  order 
th.it  W may  be  equal  to  144P.  Ans,  48. 

52.  The  distance  between  two  consecutive  threads  of  a 
screw  is  a  quarter  of  an  inch,  and  the  length  of  the  power 
arm  is  5  feet;  find  what  weight  will  be  sustained  by  a 
power  of  1  lb.  Ans.  4807T  lbs. 

53.  How  many  turns  must  be  given  to  a  screw  formed 
upon  u  cylinder  whose  length  is  10  ins.,  and  circumference 
5  ins.,  that  a  power  of  2  ozs.  may  overcome  a  pressure  of 

100  ozs.  ?  Ans.  100. 

54.  A  screw  is  made  to  revolve  by  a  force  of  2  lbs. 
applied  at  the  end  of  a  lever  3.5  ft.  long;  if  the  distance 
between  the  threads  be  J  in.,  what  pressure  can  be  pro- 
duced ?  Ans.  9  cwts.  1  qr.  20  lbs. 

55.  The  length  of  the  power-arm  is  15  inches ;  find  the 
distance  between  two  consecutive  threads  of  the  screw, 
that  the  mechanical  advantage  may  be  30.      Ans.  n  ins. 

56.  A  weight  of  W  pounds  is  suspended  from  the  block 
of  a  single  movable  pulley,  and .  the  end  of  the  cord  in 
which  the  power  acts,  is  fastened  at  the  distance  of  b  ft. 
from  the  fulcrum  of  a  horizontal  lever,  a  ft,  long,  of  the 
second  kind  ;  find  the  force,  P,  which  must  be  applied  per- 
pendicularly at  the  extremity  of  the  lever  to  sustain  W. 

Ans.  P  =  — -• 
2a 

57.  In  a  steelyard,  the  weight  of  the  beam  is  10  lbs.,  and 
the  distance  of  its  centre  of  gravity  from  the  fulcrum  is 
2  ins.,  find  where  a  weight  of  4  lbs.  must  be  placed  to  bal- 
ance it.  Ans.  At  5  ins. 


EXAMPLES.  215 

58.  A  body  whose  weight  is  V%  lbs.,  is  placed  on  a  rough 
plane  inclined  to  the  horizon  at  an  angle  of  45°.  The  co- 
efficient of  friction  being — =J  find  in  what  direction  a  force 

V3 

of  (V3  —  1)  lbs.  must  act  on  the  body  in  order  just  to 
support  it.  Aus.  At  an  angle  of  30*  to  the  plane. 

59.  A  rough  plane  is  inclined  to  the  horizon  at  an  angle 
oi  60°  ;  find  the  magnitude  and  the  direction  of  the  least 
force  which  will  prevent  a  body  weighing  100  lbs.  from  slid- 

*ng  down  the  plane,  the  coefficient  of  friction  being  — —  • 

V  3 

Ans.  50  lbs.  inclined  at  30°  to  the  plane. 


CHAPTER    VIII. 

THE  FUNICULAR*  POLYGON— THE  CATENARY 
ATTRACTION. 

130.  Equilibrium  of  the  Funicular  Polygon. — If  a 

cord  whose  weight  is  neglected,  is  suspended  from  two  fixed 
points,  A  and  B,  and  if  a  series  of  weights,  P19  P2,  P3, 
etc.,  be  suspended  from  the  given  points  Qt.  Q2,  Q3,  etc., 
the  cord  will,  when  in  equilibrium,  form  a  polygon  in  a 
vertical  plane,  which  is  called  the  Funicular  Polygon. 

Let  the  tensions  along 
the  successive  portions 
of  the  cord,  AQX,  QtQ2, 
Q2Q3,  etc.,  be  respec- 
tively Tl9  T2i  Tz,  etc., 
and  let  01?  02,  03,  etc., 
be  the  inclinations  of 
these  portions  to  the 
horizon.      Then    Q1   is 

in  equilibrium  under  the  action  of  three  forces  viz.,  Pu 
acting  vertically,  T19  the  tension  of  the  cord  AQ19  and  T2, 
the  tension  of  Qt  Q2.     Resolving  these  forces  we  have, 

for  horizontal  forces,        Tx  cos  Bt  —  T2  cos  62  =  0,  (1) 

for  vertical  forces,  Pt  +  T2  sin  62  —  Tt  sin  (?,  =  0,  {%) 

In  the  same  way  for  the  point  Q2  we  have, 

for  horizontal  forces,         T2  cos  02  —  Tz  cos  03  =  0,  (3) 

for  vertical  forces,  P2  +  T3  sin  03  —  T2  sin  02  =  0,  (4) 

*  The  term,  Funicular,  hae  reference  alone  to  the  cord,  and  has  no  mechanical 
significance. 


EQUILIBRIUM   OF  THE  FUNICULAR   POLYGON.       217 

Hence  from  (1)  and  (3)  we  have 

Tt  cos  0j  =  T2  cos  02  =  Tz  cos  03  =  etc. , 

that  is,  the  horizontal  components  of  the  tensions  in  the  dif- 
ferent portions  of  the  cord  are  constant.  Let  this  constant 
be  denoted  by  T;  then  we  have 


T                        T                        T 

rp                 ■*-        .      rp                           •       T     • 

etc., 

1    "  COS   Bx  '           2   ~~  COS  02  '           3   ~  COS  0S  ' 

which  in  (2)  and  (4)  give 

Px  +  Ttan  02  —  Ttan  0t  =  0, 

(5) 

P2  +  7*  tan  03  —  Ttan  02  =  0, 

(6) 

and  from  (5)  and  (6)  we  have 

p 

tan  0j  =  tan  02  +  -^, 


p 

and  tan  02  =  tan  03  +  -^ 


p 

Similarly         tan  63  =  tan  04  +  -~. 

and 


(7) 


P 

tan  04  =  tan  06  -f  -£. 
etc.,  etc. 


If  we  suppose  the  weights  Pl9  P2,  etc.,  each  equal  to  W, 
(7)  becomes 

tan  6X  —  tan  02  =  tan  02  —  tan  03  =  tan  03  —  tan  04 

-....-5-  (8) 

Hence,  £Ae  tangents  of  the  successive  inclinations  form  a 
series  in  Arithmetic  Progression.     In  the  figure  06  ==  0, 
10 


218      CONSTRUCTION   OF  THE  FUNICULAR   POLYGON. 


W 

tan  04  =  -™  ;     tan  03 
tan  02  =  -=- ;  tan  01 


2W 
T  ; 

4JF 
T  ' 


(9) 


etc. 


131.  To  Construct  the  Funicular  Polygon  when 
the  Horizontal  Projections  of  the  successive  Por- 
tions of  the  Cord  are  all  equal.— Let  Q5Q^,  04ff3,  q3q2-> 
q9'ji>  etc.,  be  all  of  constant  length  =  «,  and  let  Q3q3  =  c. 
Then  since  by  (9)  of  Art. 
130,  the  tangents  of  04,  03, 
02,  01?  etc.,  are  as  1,  2,  3, 
4,  etc.,  we  have 

Q2m  =  2Qzqz  =  2c; 

Qxn  =  3 $3^3  =  dc;  etc. 


A 


•JOE 


Fig.70 


q3  q*  q> 


Hence,  taking  the  middle  point,  0,  of  the  horizontal 
portion,  (?5(?4,  as  origin,  and  the  horizontal  and  vertical 
lines  through  it  as  axes  of  x  and  y,  the  co-ordinates  of  Q3 
are  (fa,  c) ;  those  of  Q2  are  (fa,  3c) ;  those  of  Qt  are  (fa, 
6c),  and  those  of  the  nth  vertex  from  §4  are  evidently 


2n  +  1 

x  =  — - —  •  a ; 


y 


n  (n  +  1) 

2     ~ 


c. 


Eliminating  n  from  these  equations  we  get 

»2 


a*  = 


_  2a2y       a* 
~V  +  4 


(1) 


which,  being  independent  of  w,  is  satisfied  by  all  the  ver- 
tices indifferently,  and  is  therefore  the  equation  of  a  curve 
passing  through  all  the  vertices  of  the  polygon,  and 
denotes  a  parabola  whose  axis  is  the  vertical  line,  OY,  and 


8 


whose  vertex  is  vertically  below  0  at  a  distance 

The  shorter  the  distances  ()4  Q3,  QiQ2,  etc.,  the  more 
nearly  does  the  funicular  polygon  coincide  with  the  para- 
bolic curve. 


CORD   SUPPORTING    LOAD. 


210 


132.  Cord  Supporting  a  Load  Uniformly  Dis- 
tributed over  the  Horizontal. — If  the  number  of  vertices 
of  the  polygon  be  very  great,  and  the  suspended  weights  all 
equal  so  that  the  load  is  distributed  uniformly  along  the 
straight  line,  FE,  the  parabola  which  passes  through  all  the 
vertices,  virtually  coincides  with  the  cord  or  chain  forming 
the  polygon,  and  gives  the  figure  of  the  Suspension  Bridge. 
In  this  bridge  the  weights  suspended  from  the  successive 
portions  of  the  chain  are  the  weights  of  equal  portions  of 
the  flooring.  The  weight  of  the  chain  itself  and  the 
weights  of  the  sustaining  bars  are  neglected  in  comparison 
with  the  weight  of  flooring  and  the  load  which  it  carries. 


Fig.7l 


Let  the  span,  AB,  =  2a,  and  the  height,  OD,  =  h. 
Then  the  equation  of  the  parabola  referred  to  the  vertical 
and  horizontal  axes  of  x  and  y,  respectively,  through  0,  is 


y2  =  kmx, 


(i) 


4ra  being  the  parameter. 

Because  the  load  between  0  and  A  is  uniformly  dis- 
tributed over  the  horizontal,  OE,  its  resultant  bisects  OE 
at  0;  therefore  the  tangents  at  A  and  0  intersect  at  G 
(Art.  62). 

From  (1)  we  have 


dy 
dx 


2m 

y 


3L 

2xy 


220  CORD   SUPPORTING   LOAD. 

which  is  the  tangent  of  the  inclination  of  the  curve  at  any 
point  {x,  y)  to  the  axis  of  x.  Hence  the  tangent  at  the 
point  of  support,  A,  makes  with  the  horizon  an  angle,  «, 

2h 
whose  tangent  is  — ,  which  also  is  evident  from  the  tri- 
angle ACE. 

Let  W  be  the  weight  on  the  cord  ;  then  £  W  is  the  weight 
on  OA,  and  therefore  is  the  vertical  tension,  V,  at  A.  Then 
the  three  forces  at  A  are  the  vertical  tension  V  =  \W,  the 
total  tension  at  the  end  of  the  cord,  acting  along  the 
tangent  AC,  and  the  horizontal  tension,  T,  which  is  every- 
where the  same  (Art.  130).  Hence,  by  the  triangle  of 
forces  (Art.  31)  these  forces  will  be  represented  by  the 
three  lines,  AE,  AC,  CE,  to  which  their  directions  are 
respectively  parallel ;  therefore  we  have  for  the  horizontal 
tension 

T  =  AE  cot  a  =  W%t9 
4A 

and  the  total  tension  at  A  is 


, w     , 

4ft 


EXAM  PLE, 

The  entire  load  on  the  cord  in  (Fig.  71)  is  320000  lbs.; 
the  span  is  150  ft.  and  the  height  is  15  ft. ;  find  the  tension 
at  the  points  of  support  and  at  the  lowest  point  and  also  the 
inclination  of  the  curve  to  the  horizon  at  the  points  of 
support. 

tan  a  =  —  =  .4 ;      .'.     a  =  21°  48'. 

a  9 

The  vertical  tension  at  each  point  of  support  is 
V=  i  weight  =  160000  lbs.; 


THE   COMMON    CATENARY 

the  horizontal  tension  is 

T=  W  "-  =  400000  lbs.; 
4/i  ' 

and  the  total  tension  at  one  end  is 


221 


VV2+  T2  =  430813  lbs. 

133.    The   Common  Catenary.— Its    Equation.— A 

catenary  is  the  curve  assumed  by  a  perfectly  flexible  cord 
when  its  ends  are  fastened  at  two  points,  A  and  B,  nearer 
together  than  the  length  of  the  cord.  When  the  cord  is  of 
constant  thickness  and  density,  i.  e.,  when  equal  portions  of 
it  are  equally  heavy,  the  curve  is  called  the  Common 
Catenary,  which  is  the  only  one  we  shall  consider. 

Let  A  and  B  be  the  fixed 
points  to  which  the  ends  of 
the  cord  are  attached ;  the 
cord  will  rest  in  a  vertical 
plane  passing  through  A  and 
B,  which  may  be  taken  to  be 
the  plane  of  the  paper  Let 
C  be  the  lowest  point  of  the 
catenary;  take  this  as  the 
origin  of  co-ordinates,  and 
let  the  horizontal  line 
through  O  be  taken  for  the 
axis  of  x,  and  the  vertical 
line  through  C  for  the  axis  of  y.  Let  (x,  y)  be  any  point, 
P,  in  the  curve  ;  denote  the  length  of  the  arc,  CP,  by  s  ; 
let  c*  be  the  length  of  the  cord  whose  weight  is  equal  to 
the  tension  at  C  ;  and  T  the  length  of  the  cord  whose 
weight  is  equal  to  the  tension  at  P. 


*  The  weight  of  a  unit  of  length  of  the  cord  being  here  taken  as  the  unit  of 
weight. 


222  THE    COMMON   CATENARY. 

Then  the  arc,  CP,  after  it  has  assumed  its  permanent 
form  of  equilibrium,  may  be  considered  as  a  rigid  body 
kept  at  rest  by  three  forces,  viz.:  (1)  T,  the  tension,  acting 
at  P  along  the  tangent,  (2)  c,  the  horizontal  tension  at  the 
lowest  point  C,  and  (3)  the  weight  of  the  cord,  CP,  acting 
vertically  downward,  and  denoted  by  s.  Draw  PT'  the 
tangent  at  P,  meeting  the  axis  of  y  at  T'.  Then  by  the 
triangle  of  forces  (Art.  31),  these  forces  may  be  represented 
by  the  three  lines  PT',  NP,  T'N,  to  which  their  directions 
are  respectively  parallel.     Therefore 

T'N  _  weight  of  CP 
NP  '  '   tension  at  G  * 

dy       s  ,  s 

Differentiating,  substituting  the  value  of  ds,  and  reducing, 
we  have 

fdy\ 

dx 


Integrating,  and  remembering  that  when  x  =  0,  ~y-  =  0, 
we  obtain 

where  e  is  the  Naperian  base.    Solving  this  equation  for 

~,  we  obtain 
dx 


THE   COMMON  CATENARY.  %2'S 

and   by  integration,  observing  that  y  =  0   when   x  =  0, 
we  have 


c  (  x-        -x\ 
V  =  g \  ec  +  e  c  I  —  <?, 


(2) 


which  is  the  equation  required.  We  may  simplify  this 
equation  by  moving  the  origin  to  the  point,  0,  at  a  dis- 
tance equal  to  c  below  C,  by  putting  y  —  c  for  y,  so  that 
(2)  becomes, 


-$(*  +  •"*)• 


y  =  z\f  +e"),  (3) 

which  is  the  equation  of  the  catenary,  in  the  usual  form. 
The  horizontal  line  through  0  is  called  the  directrix*  of 
the  catenary,  and  0  is  called  the  origin. 

Cok.  1. — To  find  the  length  of  the  arc,  CP,  we  have 


1  +  ^dx 


=  yi  +  l\f  -  e  c)  dx,  from  (1), 
=zi(&  +e~t)dz;  (4) 

the  constant  being  =  0,  since  when  x  =  0,  5  =  0. 

This  equation  may  also  be  found  immediately  by  equa* 

ting  the  values  of  ■—  in  (a)  and  (1). 

*  See  Price's  Anal.  Mechs.,  Vol.  I,  p.  21«. 


224  THE   COMMON   CATENARY. 

Cor.  2. — Since  c  =  00  is  the  length  of  the  cord  whose 
weight  is  equal  to  the  tension  of  the  curve  at  the  lowest 
point,  C,  it  follows  that,  if  the  half,  BO,  of  the  curve  were 
removed,  and  a  cord  of  length  c,  and  of  the  same  thickness 
and  density  as  the  cord  of  the  curve,  were  joined  to  the 
arc  GP,  and  suspended  over  a  smooth  peg  at  0,  the  curve 
would  be  in  equilibrium. 

Cor.  3. — We  have  from  the  triangle,  PNT', 

tension  at  P  _  PT 
tension  at  G  ~~  PN9 

or  7  =  ix =  f from  (3)  and  (4)' 

that  is,  the  tension  at  any  point  of  the  catenary  is  equal  to 
the  weight  of  a  portion  of  the  cord  ivhose  length  is  equal  to 
the  ordinate  at  that  point. 

Therefore  if  a  cord  of  constant  thickness  and  density 
ziangs  freely  over  any  two  smooth  pegs,  the  vertical  por- 
tions which  hang  over  the  pegs,  must  each  terminate  on 
the  directrix  of  the  catenary. 

Cor.  4. — From  (3)  and  (5)  we  have 

f  =  s*  +  *,  (6) 

and  from  (6)  we  have 

At  the  point,  P,  draw  the  ordinate,  PM,  and  from  M, 
the  foot  of  the  ordinate,  draw  the  perpendicular  MT,  Then 

PT=  ycos  MPT=  yC^, 


THE   COMMON  CATENARY.  225 

which  in  (7)  gives 

Pf  =  s  =  the  arc,  CP,  (8) 

and  since  f  =  PT2  +  TM2,  we  have  from  (6)  and  (8) 

TM  =  *.  (9) 

Therefore  the  point,  T,  is  on  the  involute  of  the  catenary 
which  originates  from  the  curve  at  C,  TM  is  a  tangent  to 
this  involute,  and  TP,  the  tangent  to  the  catenary,  is 
normal  to  the  involute,  (See  Calculus,  Art.  124).  As  TM 
is  the  tangent  to  this  last  curve,  and  is  equal  to  the  con- 
stant quantity,  c,  the  involute  is  the  equitangential  curve, 
or  tractrix  (See  Calculus,  p.  357). 

By  means  of  (8)  and  (9)  we  may  construct  the  origin  and 
directrix  of  the  catenary  as  follows  :  On  the  tangent  at  any 
point,  P,  measure  off  a  length,  PT,  equal  to  the  arc,  CP ; 
at  T  erect  a  perpendicular,  TM,  to  the  tangent  meeting  the 
ordinate  of  P  at  M ;  then  the  horizontal  line  through  M  is 
the  directrix,  and  its  intersection  with  the  axis  of  the  curve 
is  the  origin. 

Cor.  5. — Combining  (2)  and  (5)  we  obtain 

(y  +  c)2  =  s2  +  c8, 

therefore  s2  =  y2  +  %cy.  (10) 

The  catenary  possesses  many  interesting  geometric  and 
mechanical  properties,  but  a  discussion  of  them  would 
carry  us  beyond  the  limits  of  this  treatise.  The  student 
who  wishes  to  pursue  the  subject  further,  is  referred  to 
Price's  Anal.  Mechs.,  Vol.  I,  and  Minchin's  Statics. 


226  SPHERICAL   SHELL. 

133a.  Attraction  of  a  Spherical  Shell. — By  the  law 

of  universal  gravitation  every  particle  of  matter  attracts 
every  other  particle  with  a  force  that  varies  directly  as  the 
mass  of  the  attracting  particle,  and  inversely  as  the  square 
of  the  distance  between  the  particles. 

To  find  the  resultant  attraction  of  a  spherical  shell  of 
uniform  density  and  small  uniform  thickness,  on  a  par- 
ticle. 

(1)  Suppose  the  particle,  P, 
on  which  the  value  of  the 
attraction  is  required,  to  be 
outside  the  shell. 

Let  p  and  k  be  the  density 
and  thickness  of  the  shell,  0 
its  centre,  and  M  any  particle  of  it.  Let  OM  =  a, 
PM  =  r,  OP  =  c,  the  angle  MOP  =  0,  <p  the  angle  which 
the  plane  MOP  makes  with  a  fixed  plane  through  OP. 

Then  the  mass  of  the  element  at  M  (Art.  88)  is 
pk  a2  sin  0  dd  dtp.  The  attraction  of  the  whole  shell  acts 
along  OP;  the  attraction  of  the  elementary  mass  at  M  on 
P  in  the  direction  PM 

pk  a2  sin  6  dd  d<p 


therefore  the  attraction  of  M  on  P,  resolved  along  0P9 

_  pk  a2  sin  0  dd  d(p  c  —  a  cos  0 
r2  r 

We  shall  eliminate  0  from  this  equation  by  means  of 
/■a  =  a2  +  c2  —  2ac  cos  $ ; 
.•.    rdr  =  ac  sin  6  dd: 


(i) 


SPHERICAL   SHELL.  227 

.'.    sin  Odd  = , 

ac 

7*^  -4~  C^  —  cfi 

and  c  —  a  cos  0  =  — -~ ~ ; 

2c 

substituting  these  values  in  (1),  the  attraction  of  M  on  P 
along  PO 

To  obtain  the  resultant  attraction  of  the  whole  shell,  we 
take  the  0-integral  between  the  limits  0  and  2rr,  and  the 
y-integral  between  c  —  a  and  c  +  a. 

Hence  the  resultant  attraction  of  the  shell  on  P  along  PO 

_  4rrp^a2  _  mass  of  the  shell  .  . 

=  -*-  -        -?  w 

Since  c  is  the  distance  of  the  point  P  from  the  centre  this 
shows  that  the  attraction  of  the  shell  on  the  particle  at  P 
is  the  same  as  if  the  mass  of  the  shell  were  condensed  into 
its  centre. 

It  follows  from  this  that  a  sphere  which  is  either  homo- 
geneous or  consists  of  concentric  spherical  shells  of  uniform 
density,  attracts  the  particle  at  P  in  the  same  manner  as  if 
the  whole  mass  were  collected  at  its  centre. 

(2)  Let  the  particle,  P,  be  inside  the  sphere.  Then  we 
proceed  exactly  as  before,  and  obtain  equation  (2),  which  is 
true  whether  the  particle  be  outside  or  inside  the  sphere  • 


228  EXAMPLES. 

but  the  r-limits  in  this  case  are  a  —  c  and  a  +  c.    Hence 
from  (2)  we  have,  by  performing  the  ^-integration, 


attraction  of  shell 


(2c  -  2c)  =  0. 


therefore  a  particle  within  the  shell  is  equally  attracted  in 
every  direction,  i.  e.,  is  not  attracted  at  all. 

Cor. — If  a  particle  be  inside  a  homogenous  sphere  at  the 
distance  r  from  its  centre,  all  that  portion  of  the  sphere 
which  is  at  a  greater  distance  from  the  centre  than  the 
particle  produces  no  effect  on  the  particle,  while  the  re- 
mainder of  the  sphere  attracts  the  particle  in  the  same 
manner  as  if  the  mass  of  the  remainder  were  all  collected 
at  the  centre  of  the  sphere.  Thns  the  attraction  of  the 
sphere  on  the  particle 

_  ^ npr3  Anpr 

Hence,  within  a  homogeneous  sphere  the  attraction  varies 
as  the  distance  from  the  centre. 

The  propositions  respecting  the  attraction  of  a  uniform 
spherical  shell  on  an  external  or  internal  particle  were 
given  by  Newton  (Principia,  Lib.  I,  Prop.  70,  71).  (See 
Todhnnter's  Statics,  p.  275,  also  Pratt's  Mechs.,  p.  137, 
Price's  Anal.  Mechs.,  Vol.  I,  p.  266,  Minchin's  Statics, 
p.  403). 

EXAM  PLE  S. 

1.  The  span  AB  =800  feet,  and  CO  —  1600  feet,  find 
the  length   of  the   curve,   CA,  the   height,   CHt  and   the 


EXAMPLES.  229 

inclination,  0,  of  the  curve  to  the  horizon  at  either  point  of 
suspension. 

(1)  Here  -  =  J,   and  e  =  2-71828, 

6 

therefore  *>  =  (2-71828)*  =  1-2840, 

J? 
and  e  c  =  (2-71828)  f  =  0-7788. 

Substituting  these  values  in  (5)  we  get 

&=  800  x  0-5052  =  404-16. 

Hence  C4  =  404-16  feet. 

(2)  C7J5T  =  y  -  c  =  Ue*  +  *"*)  -  • 

=  800  x  2-062S  —  1600 
=  50- 24  feet. 

(3)  tan  0  =  ^  =*  i  (c*  -  «r*),  from  (1), 

=  0-2526, 
therefore  6  =  14°  11'. 

*  s  404-16 

Otherwise  tan  0  =  -,  from  (a),  =  =  0-2526,  as 

c  lbOO 

before. 

2.  The  entire  load  on  the  cord  in  Fig.  71  is  160000  lbs., 
the  span  is  192  ft.,  and  the  height  is  15  ft.;  find  the  tension 
at  the  points  of  support,  and  also  the  tension  at  the  lowest 
point.  Ans.  Tension  at  one  end  =  268208  lbs. 

Horizontal  tension  =  256000  " 


230  EXAMPLES. 

3.  A  chain,  ACB,  10  feet  long,  and  weighing  30  lbs.,  is 
suspended  so  that  the  height,  CH,  —  4  feet ;  find  the 
horizontal  tension,  and  the  inclination,  0,  of  the  chain  to 
the  horizon  at  the  points  of  support. 

Ans.  Horizontal  tension  =  3f  lbs.,  6  =  77°  19'. 

4.  A  chain  110  ft.  long  is  suspended  from  two  points  in 
the  same  horizontal  plane,  108  ft.  apart;  show  that  the 
tension  at  the  lowest  point  is  1.477  times  the  weight  of  the 
chain  nearly. 


PART    II. 

KINEMATICS   (MOTION), 


CHAPTER    I. 

RECTILINEAR    MOTION. 

134.  Definitions. — Velocity. — Kinematics  is  that 
branch  of  Dynamics  which  treats  of  motion  without  refer- 
ence to  the  bodies  moved  or  the  forces  producing  the  mo- 
tion (Art.  1).  Although  we  do  not  know  motion  as  free 
from  force  or  from  the  matter  that  is  moved,  yet  there  are 
cases  in  which  it  is  advantageous  to  separate  the  ideas  of 
force,  matter,  and  motion,  and  to  study  motion  in  the 
abstract,  i.  e.,  without  any  reference  to  what  is  moving,  or 
the  cause  of  motion.  To  the  study  of  pure  motion,  then, 
we  devote  this  and  the  following  chapter. 

The  velocity  of  a  particle  has  been  defined  to  be  its  rate 
of  motion  (Art.  6).  The  formula  for  uniform  and  variable 
velocities  are  those  which  were  deduced  in  Art.  7.  From 
(1)  and  (2)  of  that  Art.  we  have 

v  =  f ;  (i) 

v  =  m>  & 

in  which  v  is  the  velocity,  s  the  space,  and  t  the  time. 


232  EXAMPLES. 


EXAMPLES. 

1.  A  body  moves  at  the  rate  of  754  yards  per  hour.  Find 
the  velocity  in  feet  per  second. 

Since  the  velocity  is  uniform  we  use  (1),  hence 

v  =  -  =  ^r —  =  0.628  ft.  per  sec,  Ans. 

2.  Find  the  position  of  a  particle  at  a  given  time,  t, 
when  the  velocity  varies  as  the  distance  from  a  given  point 
on  the  rectilinear  path. 

Here  the  velocity  being  variable  we  have  from  (2) 

ds       ,  -^ 

"  =  *  =  hs> 

where  h  is  a  constant ; 

ds 
therefore      —  =  hdt ;    .•.    log  s  =  ht  +  c,  (1) 

s 

where  c  is  an  arbitrary  constant. 

Now  if  we  suppose  that  s0  is  the  distance  of  the  particle 
from  the  given  point  when  t  =  0  we  have  c  =  log  s0, 
which  in  (1)  gives 

log  —  =  kt;    or    s  =  s0e**. 

3.  A  railway  train  travels  at  the  rate  of  40  miles  per 
hour  ;  find  its  velocity  in  feet  per  second. 

A ns.  58. 66  ft. ,  per  second. 

4.  A  train  takes  7  h.  31  m.  to  travel  200  miles  ;  find  its 
velocity.  Ans.  39.02  ft.  per  sec. 

5.  If  s  =  Uz,  find  the  velocity  at  the  end  of  five  seconds. 

Ans.  300  ft.  per  sec. 

6.  Find  the  position  of  the  particle  in  Ex.  2,  when  the 
velocity  varies  as  the  time.  Ans.  s  =  sn  +  \ht\ 


ACCELERATION  ZERO.  233 

7.  Find  the  distance  the  particle  will  move  in  one 
minute,  when  the  velocity  is  10  ft.  at  the  end  of  one 
second  and  varies  as  the  time.  Ans.  18000  ft. 

135.  Acceleration. — Acceleration  has  heen  defined  to 
be  the  rate  of  change  of  velocity  (Art.  8).  It  is  a  velocity 
increment.  The  formulae  for  acceleration  are  from  (1),  (2), 
and  (3)  of  (Art.  9), 

r-fi  (i) 

/-*'  <*> 

(1)   being  for  uniform,  and    (2)    and    (3)    for  variable, 
acceleration. 

If  the  velocity  decreases,  f  is  negative,  and  (2)  and  (3) 
become 

dv  _         f     dh^  _ 

dt  ~       T;    dP  ""      /; 

and  the  velocity  and  time  are  inverse  functions  of  each 
other. 

136.  The  Relation  between  the  Space  and  Time 
when  the  Acceleration  =  0. 

Here  we  have 

d^  -  ' 

so  that  if  v0  is  the  constant  velocity  we  have 


dt  ~  "°  ' 

8  =  V0t  +  8Al 


234  ACCELERATION  CONSTANT. 

in  which  s0  is  the  space  which  the  body  has  passed  over 
when  t  =  0.  If  t  is  computed  from  the  time  the  body 
starts  from  rest,  then  s  =  v0t.  The  student  will  observe 
that  this  is  a  case  of  uniform  velocity. 

137.  The  Relation  (1)  between  the  Space  and 
Time,  and  (2)  between  the  Space  and  Velocity, 
when  the  Acceleration  is  Constant. 

(1)  Let  A  be  the  initial  position  of  d  a  p  * 
the   particle   supposed   to    be    moving  FFs-73 

toward  the  right,  P  its  position  at  any  time,  t,  from  A,  v 
its  velocity  at  that  time,  and  /  the  constant  acceleration  of 
its  velocity.  Take  any  fixed  point,  0,  in  the  line  of  motion 
as  origin,  and  let  OA  =  s0 ;  OP  =  s.  Then  the  equation 
of  motion  is 

£=/;  •  (i) 

Suppose  the  velocity  of  the  particle,  at  the  point  A  to  be 
r0,  then  when  t  =  0,  v  =  v9f  hence  c  =  v0,  and 

ds 

But  when  t  =  0,  s  =  s0;  hence  c'  =  s0,  and 

s  =  tffi  +  V  +  s0,  (3) 

Hence  if  a  particle  moves  from  rest  from  the  origin  0,  with 
a  constant  acceleration,  we  have 

*  Called  initial  velocity  and  space  respectively,  or  the  velocity  the  particle  haa 
and  space  it  has  moved  over  at  the  instant  t  oegins  to  be  reckoned. 


ACCELERATION    VARIABLE.  235 

s  =  if?,  a) 

and  thus  the  space  described  varies  as  the  square  of  the 
time. 

(2)  From  (1)  we  have 

...    g  =  2/*  +  0. 

But  when  s  =  s0,  v  =  v0 ;  hence  C  =  v02  —  2/s0,  and 
therefore 

v*  =  2fs  +  v02  -  2/V  (5) 

Equations  (2)  and  (3)  give  the  velocity  "and  position  of  the 
particle  in  terms  of  I ;  and  (5)  gives  the  velocity  in  terms 
of  s. 

138.  When  the  Acceleration  Varies  directly  as 
the  Time  from  a  State  of  Rest,  find  the  Velocity 
and  Space  at  the  end  of  the  Time  t. 

Here  <S>  =  at> 

•"*    Jt  =  ^   +v» 

where  vQ  is  the  initial  velocity  ; 

.-.    s  =  %at5  +  v0t, 
the  initial  space  being  0  since  t  is  estimated  from  rest. 

139.  When  the  Acceleration  Varies  directly  as 
the  Distance  from  a  given  Point  in  the  line  of  Mo- 
tion, and  is  negative,  find  the  Relation  between 
the  Space  and  Time. 


236 

EXAMPLES. 

Here 

.-.    2dsd^=-2itda; 

dp 

£=*w-4, 

by  calling'  s 

0  the  value  of  s  when  the  particle  is  at  rest. 

.-.  -  ,  ds       M 

Vs02  -  s2 


the  negative  sign  being  taken    since  the  particle  is  moving 
towards  the  origin  ] 


cos-*  —  =  &**, 
So 


:t'  s  =  S0  when  t  =  0 j 


<s°  cos 


#*. 


EXAM  PLES, 

1.  A  body  commences  to  move  with  d  velocity  of  30  ft. 
per  sec,  and  its  velocity  is  increased  m  each  second  by 
10  ft.     Find  the  space  described  in  5  seconds. 

Here  /  =  10,  v0  =  30,  s0  =  0,  and  t  =  5,  therefore 
from  (3)  we  have 

s  =  J  •  10  •  25  +  30  •  5  =  275,  Ans. 

2.  A  body  starting  with  a  velocity  of  10  ft.  per  sec.,  and 
moving  with  a  constant  acceleration,  describes  90  ft.  in 
4  sees. ;  find  the  acceleration.  Ans.  6J  ft.  per  sec. 

3.  Find  the  velocity  of  a  body  which  starting  from  rest 
with  an  acceleration  of  10  ft.  per  sec,  has  described  a  space 
of  20  ft.  Ans.  20  ft. 


FALLING    BODIES.  23? 

4.  Through  what  space  must  a  body  pass  under  an  accel- 
eration of  5  ft.  per  sec,  so  that  its  velocity  may  increase 
from  10  ft.  to  20  ft.  per  sec.  ?  Ans.  30  ft. 

5.  In  what  time  will  a  body  moving*  with  an  accelera- 
tion of  25  ft.  per  sec,  acquire  a  velocity  of  1000  ft.  per 
second?  Ans.  40  sees. 

6.  A  body  starting  from  rest  has  been  moving  for  5  min- 
utes, and  has  acquired  a  velocity  of  30  miles  an  hour; 
what  is  the  acceleration  in  feet  per  second  ? 

Ans.  %\  ft.  per  sec. 

7.  If  a  body  moves  from  rest  with  an  acceleration  of  §  ft. 
per  sec,  how  long  must  it  move  to  acquire  a  velocity  of 
40  miles  an  hour?  Ans.  88  sees. 

140.  Equations  of  Motion  for  Falling  Bodies. — 

The  most  important  case  of  the  motion  of  a  particle  with  a 
constant  acceleration  in  its  line  of  motion  is  that  of  a  body 
moving  under  the  action  of  gravity,  which  for  small  dis- 
tances above  the  earth's  surface  may  be  considered  constant. 
When  a  body  is  allowed  to  fall  freely,  it  is  found  to  acquire 
a  velocity  of  about  32.2  feet  per  second  during  every  second 
of  its  motion,  so  that  it  moves  with  an  acceleration  of  32.2 
feet  per  second  (Art.  21).  This  acceleration  is  less  at  the 
summit  of  a  high  mountain  than  near  the  surface  of  the 
earth  ;  and  less  at  the  equator  than  in  the  neighborhood  of 
the  poles ;  i.  e.,  the  velocity  which  a  body  acquires  in  falling 
freely  for  one  second  varies  with  the  latitude  of  the  place, 
and  with  its  altitude  above  the  sea  level ;  but  is  independ- 
ent of  the  size  of  the  body  and  of  its  mass.  Practically, 
however,  bodies  do  not  fall  freely,  as  the  resistance  of  the 
air  opposes  their  motion,  and  therefore  in  practical  cases  at 
high  speed  (e.  g.,  in  artillery)  the  resistance  of  the  air  must 
be  taken  into  account.     But  at  present  we  shall  neglect 

*  in  each  case  the  hody  is  supposed  to  start  from  re6t  unless  otherwise  stated. 


238  FALLING    BODIES. 

this  resistance,  and  consider  the  bodies  as  moving  in  vacuo 
under  the  action  of  gravity,  i.  e.,  with  a  constant  accelera- 
tion of  about  32.2  feet  per  second. 

As  neither  the  substance  of  the  body  nor  the  cause  of 
the  motion  needs  to  be  taken  into  consideration,  all  prob- 
lems relating  to  falling  bodies  may  be  regarded  as  cases  of 
accelerated  motion,  and  treated  from  purely  geometric 
considerations.  Therefore  if  we  denote  the  acceleration  by 
g,  as  in  Art.  23,  and  consider  the  particle  in  Art.  137  to  be 
moving  vertically  downwards,  then  (2),  (3),  (5)  of  Art.  137 
become,  by  substituting  #  for/, 

v  =  gt  +  v0,  \ 

8  =  igt2  +  v0t  +  s0,    |  (A) 

t?  =  2gs  +  V  -  2gs0, ) 

s  being  measured  as  before  from  a  fixed  point,  0,  in  the 
line  of  motion. 

Suppose  the  particle  to  be  projected  downward  from  0, 
then  A  commences  with  0  and  sQ  =  0.  Hence  (A)  be- 
comes 

v  =  gt  -f-  v0,  (1) 

s  =  \gP  +  V,  (2) 

v*  =  2gs  +  V-  (3) 

As  a  particular  case  suppose  the  particle  to  be  dropped 
from  rest  at  0  (Fig.  73).  Then  A  coincides  with  0,  and 
s0  =  0,  vQ  =  0.     Hence  equations  (A)  become 

v  =  gt,  (4) 

s  =  \gt\  (5) 

v2  =  2gs.  (6) 


PARTICLE  PROJECTED    UPWARDS.  239 

141.  When  the  Particle  is  Projected  Vertically 
Upwards. — Here  if  we  measure  s  upwards  from  the  point 
of  projection,  0,  the  acceleration  tends  to  diminish  the 
space  and  therefore  the  acceleration  is  negative,  and  the 
equation  of  motion  is  (Art.  135) 

<Ps  _ 

In  other  respects  the  solution  is  the  same.  Taking 
therefore  s0  =  0  in  (A)  and  changing  the  sign  of  g*  we 
obtain 

v  =  vo  —  9*>  (1) 

*  =  V  -  isP>  (2) 

v8  =  V  -  %98.  (3) 

Cor.  1. — Tlie  time  during  which  a  particle  rises  when 
projected  vertically  upwards. 

When  the  particle  reaches  its  highest  point,  its  velocity 
is  zero.  If  therefore  we  put  v  =  0  in  (1),  the  correspond- 
ing value  of  t  will  be  the  time  of  the  particle  ascending  to  a 
state  of  rest. 

•  t  =  **. 

9 

Cor.  2. — The  time  of  flight  before  returning  to  the  start- 
ing point. 

From  (2)  we  have  the  distance  of  the  particle  from  the 
starting  point  after  t  seconds,  when  projected  vertically 
upwards  with  the  velocity  v0.  Now  when  the  particle  has 
risen  to  its  maximum  height  and  returned  to  the  point  of 
projection,  5  =  0.  If,  therefore,  we  put  s  =  0  in  (2),  and 
solve  for  t,  we  shall  get  the  time  of  flight.     Therefore, 

*  g  is  positive  or  negative  according  as  the  particle  is  descending  or  as 
cending. 


240  PARTICLE   PROJECTED    UPWARDS, 


2v 
which  gives  t  =  0,    or     — -■ 


The  first  value  of  t  shows  the  time  before  the  particle 

starts,  the  latter  shows  the  time  when  it  has  returned. 

2v 
Hence,  the  ivhole  time  of  flight  is  —  -,  which  is  just  double 

the  time  of  rising  (Cor.  1) ;  that  is,  the  time  of  rising  equals 
the  time  of  falling. 

The  final  velocity,  by  (1)  of  Art.  140,  =  gt  =  g  x  ^ 

(Cor.  1)  =  v0  ;  hence  a  body  returns  to  any  point  in  its 
path  with  the  same  velocity  at  which  it  left  it.  In  other 
words,  a  body  passes  each  point  in  its  path  with  the  same 
velocity,  whether  rising  or  falling,  since  the  velocity  at  any 
point  may  be  considered  as  a  velocity  of  projection. 

Cor.  3. — TJie  greatest  height  to  which  the  particle  will 
rise. 

At  the  summit  v  =  0,  and  the  corresponding  value  of  s 
will  be  the  greatest  height  to  which  the  particle  will  rise  ; 
when  v  =  0,  (3)  becomes 

v02  =  2gs ; 

.•.    s  =  ^« 

Cor.  4. — Since  v^  =  2gsf  where  s  is  the  height  from 
which  a  body  falls  to  gain  the  velocity  vQ,  it  follows  that  a 
body  will  rise  through  the  same  space  in  losing  a  velocity 
v0  as  it  would  fall  through  to  gain  it. 


EXAMPLES.  241 


EXAMPLES. 

1.  A  body  projected  vertically  downwards  with  a  velocity 
jf  20  ft.  a  sec.  from  the  top  of  a  tower,  reaches  the  ground 
in  2.5  sees.;  find  the  height  of  the  tower. 

Here  t  =  2 J,  and  v0  —  20  ;  assume  g  =  32.  Then 
from  (2)  of  Art.  140  we  have 

s  —  ia*3UL  +  20  x  f  =  150  ft. 

2.  A  body  is  projected  vertically  upwards  with  a  velocity 
of  200  ft.  per  second ;  find  the  velocity  with  which  it  will 
pass  a  point  100  ft.  above  the  point  of  projection. 

Here  v0  =  200,  s  =  100  ;  therefore  from  (3)  we  have 
&  =  40000  -  6400  =  33600 ; 
.-.     <y  =  40\/2T. 

3.  A  man  is  ascending  in  a  balloon  with  a  uniform 
velocity  of  20  ft.  per  sec,  when  he  drops  a  stone  which 
reaches  the  ground  in  4  sees.;  find  the  height  of  the 
balloon. 

Here  v0  =  20,  and  t  =  4 ;  therefore  from  (2)  we  have, 
after  changing  the  sign  of  the  second  member  to  make  the 
result  positive, 

s  =  —  (80  —  256)  =  176, 

which  was  the  height  of  the  balloon. 

4.  A  body  is  projected  upwards  with  a  velocity  of  80  ft. ; 
after  what  time  will  it  return  to  the  hand  ? 

Ans.  5  seconds. 

5.  With  what  velocity  must  a  body  be  projected  ver- 
tically upwards  that  it  may  rise  40  ft.  ? 

Ans.  16  VlO  ft.  per  sec. 
11 


242  COMPOSITIOX   OF    VELOCITIES. 

6.  A  body  projected  vertically  upwards  passes  a  certain 
point  with  a  velocity  of  80  ft.  per  sec. ;  how  much  higher 
will  it  ascend  ?  Ans.  100  ft. 

7.  Two  balls  are  dropped  from  the  top  of  a  tower,  one  of 
them  3  sees,  before  the  other  ;  how  far  will  they  be  apart 
D  sees,  after  the  first  was  let  fall  ?  Ans.  336  ft. 

8.  If  a  body  after  having  fallen  for  3  sees,  breaks  a  pane 
sf  glass  and  thereby  loses  one-third  of  its  velocity,  find  the 
entire  space  through  which  it  will  have  fallen  in  4  sees. 

Ans.  224  ft. 

142.  Composition  of  Velocities. — (1)  From  the  Par- 
allelogram of  Velocities,  (Art.  29,  Fig.  2),  we  see  that  if  AB 
represents  in  magnitude  and  direction  the  space  which 
would  be  described  in  one  second  by  a  particle  moving  with 
a  given  velocity,  and  AC  represents  in  magnitude  and 
direction  the  space  which  would  be  described  in  one  second 
by  another  particle  moving  with  its  velocity,  then  AD,  the 
diagonal  of  the  parallelogram,  represents  the  result  ai.t 
velocity  in  magnitude  and  direction. 

(2)  Hence  the  resultant  of  any  tivo  velocities,  as  AB,  BD, 
(Fig.  2),  is  a  velocity  represented  by  the  third  side,  DA.  of 
the  triangle  ABD ;  and  if  a  point  have  simultaneously, 
velocities  represented  by  AB,  BC,  CA,  the  sides  of  a  trian- 
gle, taken  in  the  same  order,  it  is  at  rest. 

The  lines  which  are  taken  to  represent  any  given  forces 
may  clearly  be  taken  to  represent  the  velocities  which 
measure  these  forces  (Art.  19),  therefore  from  the  Polygon 
and  Parallelopiped  of  Forces  the  Polygon  and  Parallel- 
opiped  of  Velocities  follow. 

(3)  Hence,  if  any  number  of  velocities  be  represented  in 
magnitude  and  direction  by  the  sides  of  a  closed  polygon, 
taken  all  in  the  same  order,  the  resultant  is  zero. 

(4)  Also,  if  three  velocities  be  represented  in  magnitude 


RESOLUTION    OF    VELOCITIES.  243 

and  direction  by  the  three  edges  of  a  fitrallelopiped,  lite  re- 
sultant velocity  will  be  represented  by  Ihe  diagonal. 

(5)  When  there  are  two  velocities  or  three  velocities  in 
two  or  in  three  rectangular  directions,  the  resultant  is  the 
square    root    of    the    sum    of    their    squares.       Thus,   if 

-r.>  ^.-j  -tt,  -r,,  are  the  velocities  of  the  moving  point  and 
df     dt     dt     dt 

its  components  parallel   to  the  axes,  we  have  from  (2)  of 

Art.  30,  

dt        V  \dl 


and  from  (1)  of  Art.  34, 


ds 

dt 


= s/<gf+  m*  ©*•     • 


143.  Resolution  of  Velocities. — As  the  diagonal  of 
the  parallelogram  (Fig.  2),  whose  sides  represent  the  com- 
ponent velocities  was  found  to  represent  the  resultant 
velocity,  so  any  velocity,  represented  by  a  given  straight 
line,  may  be  resolved  into  component  velocities  represented 
by  the  sides  of  the  parallelogram  of  which  the  given  line 
is  the  diagonal. 

It  will  be  easily  seen  that  (2)  of  Art.  134  is  equally 
applicable  whether  the  point  be  considered  as  moving  in  a 
straight  line  or  in  a  curved  line  ;  but  since  in  the  latter 
case  the  direction  of  motion  continually  changes,  the  mere 
amount  of  the  velocity  is  not  sufficient  to  describe  the 
motion  completely,  so  it  will  be  necessary  to  know  at  every 
instant  the  direction,  as  well  as  the  magnitude,  of  the  point's 
velocity.  In  such  cases  as  this  the  method  commonly  em- 
ployed, whether  we  deal  with  velocities  or  accelerations, 
consists  mainly  in  studying,  not  the  velocity  or  acceleration, 
directly,  but  its  components  parallel  to  any  three  assumed 
rectangular  axes.     If  the  particle  be  at  the  point  (x,  y,  z), 


244  EXAMPLES. 

at  the   time  t,  and   if  we   denote  its   velocities    parallel 
respectively  to  the  three  axes  by  u,  v,  w,  we  have 

dx  _  dy  _  dz 

di~Vx'   It  -Vy>  It 

Denoting  by  v  the  velocity  of  the  moving  particle  along 
the  curve  at  the  time  t,  we  have  as  above 


and  if  a,  ft  y  be  the  angles  which  the  direction  of  motion 
along  the  curve  makes  with  the  axes,  we  have,  as  in  (2)  of 

(Art.  34), 

dx       ds 

-=-  =  —  cos  a  =  v  cos  a  =  vx ; 

dt       dt 

dy      ds        n  o 

dz       ds 

-r-  =  -r;  cos  y  =  v  cos  y  =  v*. 

dt       dt 

Hen<;3  each  of  the  components  -jr,  -~,  -r,  is  to  be 

found  from  the  whole  velocity  by  resolving  the  velocity, 
i.  e.,  by  multiplying  the  velocity  by  cosine  of  the  angle 
between  the  direction  of  motion  and  that  of  the  compo- 
nent. 

EXAM  PLES, 

1.  A  body  moves  under  the  influence  of  two  velocities, 
at  right  angles  to  each  other,  equal  respectively  to  17.14  ft. 
and  13.11  ft.  per  second.  Find  the  magnitude  of  the 
resultant  motion,  and  the  angles  into  which  it  divides  the 
right  angle. 

Ans.  21.579  ft.  per  sec. ;  37°  25'  and  52°  35\ 


MOTION   ON  AN  INCLINED  PLANE.  245 

2.  A  ship  sails  due  north  at  the  rate  of  4  knots  per 
hour,  and  a  ball  is  rolled  towards  the  east,  across  her  deck, 
at  right  angles  to  her  motion  at  the  rate  of  10  ft.  per 
second.  Find  the  magnitude  and  direction  of  the  real 
motion  of  the  ball. 

Ans.  12.07  ft.  per  sec;  and  N.  56°  E. 

3.  A  boat  moves  N.  30°  E.,  at  the  rate  of  6  miles  per 
hour.     Find  its  rate  of  motion  northerly  and  easterly. 

Ans.  5.2  miles  per  hour  north,  and  3  miles  per  hour 
east. 

144.  Motion  on  an  Inclined  Plane. — By  an  exten- 
sion of  the  equations  of  Art.  140,  we  may  treat  the  case  of 
a  particle  sliding  from  rest  down  a  smooth  inclined  plane. 
As  this  is  a  very  simple  case  in  which  an  acceleration  is 
resolved,  it  is  convenient  to  treat  of  it  in  this  part  of  our 
work  ;  yet  as  it  properly  belongs  to  the  theory  of  con- 
strained motion,  we  are  unable  to  give  a  complete  solution 
of  it,  until  the  principles  of  such  motion  have  been  ex- 
plained in  a  future  chapter. 

Let  P  be  the  position  of  the  particle  at 
any  time,  t,  on  the  inclined  plane  OA,  OP 
=  s,  its  distance  from  a  fixed  point,  0,  in 
the  line  of  motion,  and  let  «  be  the  inclina- 
tion of  OA  to  the  horizontal  line  AB.  Let 
Pb  represent  g,  the  vertical  acceleration  with 
which  the  body  would  move  if  free  to  fall.  Eesolve  this 
into  two  components,  Pa  =  g  sin  «  along,  and  Pc  =  g 
cos  a  perpendicular  to  OA.  The  component  g  cos  a  pro- 
duces pressure  on  the  plane,  but  does  not  affect  the  motion. 
The  only  acceleration  down  the  plane  is  that  component  of 
the  whole  acceleration  which  is  parallel  to  the  plane,  viz., 
g  sin  a.     The  equation  of  motion,  therefore,  is 

d2s 

jp=g  sin  a,  (1) 


24 G  DESCENT  DOWN   CHORDS   OF  A    CIRCLE. 

the  solution  of  which,  as  g  sin  «  is  constant,  is  included  in 
that  of  Art.  140;  and  all  the  results  for  particles  moving 
vertically  as  given  in  Arts.  140  and  141  will  be  made  to 
apply  to  (1)  by  writing  #  sin  a  for  g.  Thus,  if  the  particle 
be  projected  down  or  up  the  plane,  we  get  from  (1),  (2),  (3) 
of  Arts.  140  and  141,  by  this  means 


v  =  v0  ±g  sin  cc*t, 

(2) 

8  =  v0t  ±  \g  sin  «•  t% 

(3) 

ip  =  v02  ±  ty  sin  a'Sy 

(4) 

in  which  the  +  or  —  sign  is  to  be  taken  according  as  the 
body  is  projected  down  or  up  the  plane. 

If  the  particle  starts  from  rest  from  0,  we  get  from  (4), 
(5),  (6)  of  Art.  140 

v  =  g  sin  cct,  (5) 

s  =  \g  sin  «•  t\  (6) 

«r>  =  2#sin«.s.  (7) 

Cor.  1. — Tlie  velocity  acquired  by  a  particle  in  falling 
down  a  given  inclined  plane. 

Draw  PC  parallel  to  AB  (Fig.  74),  then  if  v  be  the 
velocity  at  P,  we  have  from  (7) 

v8  =  2g  sin  a»8 
=  2^.00. 

Hence,  from  (6)  of  Art.  140  the  velocity  is  the  same  at  P 
as  if  the  particle  had  fallen  through  the  vertical  space  OC  ; 
that  is,  the  velocity  acquired  in  falling  down  a  smooth 
inclined  plane  is  the  same  as  would  be  acquired  in  falling 
freely  through  the  perpendicular  height  of  the  plane. 


DESCENT  DOWN   CHORDS    OF  A    CIRCLE.  24 1 

Cor.  2. —  When  the  particle  is  projected  up  the  plane  with 
a  given  velocity,  to  find  how  high  it  will  ascend,  and  the  time 
of  ascent. 

From  (4)  we  have 

v2  =  v02  —  2g  sin  «•& 

When  v  =  0  the  particle  will  stop ;  hence,  the  distance  it 
will  ascend  will  be  given  by  the  equation 

0  =  v02  —  2g  sin  a»s, 

v  2 
2g  sin  a 

To  find  the  time  we  have  from  (2) 

v  =  v0  —  g  sin  a't', 
and  the  particle  stops  when  v  =  0,  in  which  case  we  have 

g  sin  a 

From  (6)  we  derive  the  following  curious  and  useful 
result. 

145.  The  Times  of  Descent  down 
all  Chords  drawn  from  the  Highest 
Point  of  a  Vertical  Circle  are  equal. — 

Let   AB   be   the   vertical   diameter  of  the 

circle,    AC    any    cord    through    A,    a  its 

inclination  to  the  horizon  ;  join  BC  ;  then 

if  /  be  the  time  of  descent  down  AC  we  Fig>75 

have  from  (6)  of  Art.  144 

AC  =  \gt2  sin  a. 

But  AC  ss  AB  sin  a; 


248  LINE   OF  QUICKEST  DESCENT. 


or 


/2AB 


which  is  constant,  and  shows  that  the  time  of  falling  down 
any  chord  is  the  same  as  the  time  of  falling  down  the 
diameter. 

Cor. — Similarly  it  may  be  shown  that  the  times  of 
descent  down  all  chords  drawn  to  B,  the  lowest  point, 
are  equal ;  that  is,  the  time  down  CB  is  equal  to  that 
down  AB. 

146.  The  Straight  Line  of  Quickest  Descent  from 
(1)  a  Given  Point  to  a  Given  Straight  Line  (2)  from 
a  Given  Point  to  a  Given  Curve. 

(1)  Let  A  be  the  given  point  and  BO 
the  given  line.  Through  A  draw  the 
horizontal  line  AC,  meeting  CB  in  C; 
bisect  the  angle  ACB  by  CO  which  inter- 
sects in  0  the  vertical  line  drawn  through 
A  ;  from  0  draw  OP  perpendicular  to  BC; 
join  AP  ;  AP  is  the  required  line  of  quick- 
est descent. 

For  OP  is  evidently  equal  to  OA,  and  therefore  the 
circle  described  with  0  as  centre  and  with  OP  (=  OA)  for 
radius,  will  touch  the  line  BC  at  P,  and  since  the  time  of 
falling  down  all  chords  of  this  circle  from  A  is  the  same, 
AP  must  be  the  line  of  quickest  descent. 

(2)  To  find  the  straight  line  of  quickest  descent  to  a 
given  curve,  all  that  is  required  is  to  draw  a  circle  having 

.the  given  point  as  the  upper  extremity  of  its  vertical 
diameter,  and  tangent  to  the  curve.  Hence  if  DE  (Fig. 
76)  be  the  curve,  A  the  point,  draw  AH  vertical ;  and,  with 
centre  in  AH,  describe  a  circle  passing  through   A,   and 


EXAMPLES.  249 

touching  DE  at  P,  then  AP  is  the  required  line.  For,  if  we 
take  any  other  point,  Q,  in  DE,  and  draw  AQ  cutting  the 
circle  in  q,  then  the  time  down  AP  =  time  down  Aq< 
time  down  AQ.     Hence  AP  is  the  line  of  quickest  descent. 

The  problem  of  finding  the  line  of  quickest  descent  from  a  point  to' 
a  line  or  curve  is  thus  found  to  resolve  itself  into  the  purely  geometric 
problem  of  drawing  a  circle,  the  highest  point  of  which  shall  be  the 
given  point  and  which  shall  touch  the  given  line  or  curve. 

EXAMPLES.* 

1.  If  the  earth  travels  in  its  orbit  600  million  miles  in 
365J  days,  with  uniform  motion,  what  is  its  velocity  in 
miles  per  second  ?  Ans.  19-01  miles. 

2.  A  train  of  cars  moving  with  a  velocity  of  20  miles  an 
hour,  had  been  gone  3  hours  when  a  locomotive  was 
dispatched  in  pursuit,  with  a  velocity  of  25  miles  an  hour; 
in  what  time  did  the  latter  overtake  the  former  ? 

Ans.  12  hours. 

3.  A  body  moving  from  rest  with  a  uniform  acceleration 
describes  90  ft.  in  the  5th  second  of  its  motion ;  find  the 
acceleration,/,  and  velocity,  v,  after  10  seconds. 

Ans.  f  —  20;  v  =  200. 

4.  Find  the  velocity  of  a  particle  which,  moving  with  an 
acceleration  of  20  ft.  per  sec.  has  traversed  1000  ft. 

Ans.  200  ft.  per  sec. 

5.  A  body  is  observed  to  move  over  45  ft.  and  55  ft.  in 
two  successive  seconds ;  find  the  space  it  would  describe  in 
the  20th  second.  Ans.  195  ft. 

6.  The  velocity  of  a  body  increases  every  hour  at  the  rate 
of  360  yards  per  hour.  What  is  the  acceleration,/,  in  feet 
per  second,  and  what  is  the  space,  s,  described  from  rest  in 
20  seconds?  Ans.  f  =  0.3;  *  =  60  ft. 

*  In  these  examples  take  g  =  32  ft. 


250  EXAMPLES. 

7.  A  body  is  moving,  at  a  given  instant,  at  the  rate  of 
8  ft.  per  sec.;  at  the  end  of  5  sees,  its  velocity  is  19  ft.  per 
sec.  Assuming  its  acceleration  to  be  uniform,  what  was  its 
velocity  at  the  end  of  4  sees.,  and  what  will  be  its  velocity 
at  the  end  of  10  sees.  ?  Ans.  16-  8 ;  30. 

8.  A  body  is  moving  at  a  given  instant  with  a  velocity  of 
30  miles  an  hour,  and  comes  to  rest  in  11  sees.;  if  the 
retardation  is  uniform  what  was  its  velocity  5  sees,  before  it 
stopped  ?  Ans.  20  ft.  per  sec. 

9.  A  body  moves  at  the  rate  of  12  ft.  a  sec.  with  a 
uniform  acceleration  of  4 ;  (1)  state  exactly  what  is  meant 
by  the  number  4  ;  (2)  suppose  the  acceleration  to  go  on  for 
5  sees.,  and  then  to  cease,  what  distance  will  the  body 
describe  between  the  ends  of  the  5th  and  12th  sees.? 

Ans.  224  ft. 

10.  A  body,  whose  velocity  undergoes  a  uniform  retarda- 
tion of  8,  describes  in  2  sees,  a  distance  of  30  ft.;  (1)  what 
was  its  initial  velocity  ?  (2)  How  much  longer  than  the 
2  sees,  would  it  move  before  coming  to  rest  ? 

Ans.  (1)  23 ;  (2)  £  sec. 

11.  A  body  whose  motion  is  uniformly  retarded,  changes 
its  velocity  from  24  to  6  while  describing  a  distance  of  12 
ft.;  in  what  time  does  it  describe  the  12  ft.? 

A?is.  0-8  sec. 

12.  The  velocity  of  a  body,  which  is  at  first  6  ft.  a  sec, 
undergoes  a  uniform  acceleration  of  3 ;  at  the  end  of  4  sees. 
the  acceleration  ceases  ;  how  far  does  the  body  move  in  10 
sees,  from  the  beginning  of  the  motion?        Ans.  156  ft. 

13.  A  body  moves  for  a  quarter  of  an  hour  with  a  uni- 
form acceleration ;  in  the  first  5  minutes  it  describes  350 
yards  ;  in  the  second  5  minutes  420  yards ;  what  is  the 
whole  distance  described  in  a  quarter  of  an  hour? 

Ans.  1260  yds. 


EXAMPLES.  251 

14.  Two  sees,  after  a  body  is  let  fall  another  body  is 
projected  vertically  downwards  with  a  velocity  of  100  ft. 
per  sec;  when  will  it  overtake  the  former? 

Ans.  1J  sees. 

15.  A  body  is  projected  upwards  with  a  velocity  of  100 
ft.  per  sec;  find  the  whole  time  of  flight.      Ans.  %-\  sees. 

16.  A  balloon  is  rising  uniformly  with  a  velocity  of  10  ft. 
per  sec,  when  a  man  drops  from  it  a  stone  which  reaches 
the  ground  in  3  sees.;  find  the  height  of  the  balloon,  (1) 
when  the  stone  was  dropped ;  and  (2)  when  it  reached  the 
ground.  Ans.  (1)  114  ft. ;  (2)  144  1't. 

17.  A  man  is  standing  on  a  platform  which  descends 
with  a  uniform  acceleration  of  5  ft.  per  sec;  after  having 
descended  for  2  sees,  he  drops  a  ball ;  what  will  be  the 
velocny  of  the  ball  after  2  more  seconds  ?         Ans.  74  ft. 

18.  A  balloon  has  been  ascending  vertically  at  a  uniform 
rate  for  4-5  sees.,  and  a  stone  let  fall  from  it  reaches  the 
ground  in  7  sees.;  find  the  velocity,  v,  of  the  balloon  and 
the  height,  s,  from  which  the  stone  is  let  fall. 

Ans.  v  =  174f  ft.  per  sec;  s  =  784  ft.  It  the  balloon 
is  still  ascending  when  the  stone  is  let  fall  v  =  68*17  ft. 
per  sec;  s  =  306-76  ft.9 

19.  With  what  velocity  must  a  particle  be  projected 
downwards,  that  it  may  in  I  s^cs-  overtake  another  particle 
which  has  already  fallen  through  i  ft.  ? 


20.  A  person  while  ascending  in  a  balloou  with  a  vertical 

velocity  of  V  ft.  per  sec,  lets  fall  a  stone  when  he  is  h  f c. 

above  the  ground  ;  required  the  time  in  which  the  stou* 

will  reach  the  ground.  y  _i_  y  V2  4-  %gh 

Ans. ^-* 

9 


252  EXAMPLES. 

21.  A  body,  A,  is  projected  vertically  downwards  from 
the  top  of  a  tower  with  the  velocity  V,  and  one  sec.  after- 
wards another  body,  B,  is  let  fall  from  a  window  a  ft.  from 
the  top  of  the  tower  ;  in  what  time,  t,  will  A  overtake  B  ? 

,         .  2a  +  a 

Ans.  i  =  __. 

22.  A  stone  let  fall  into  a  well,  is  heard  to  strike  the 
bottom  in  t  seconds  ;  required  the  depth  of  the  well,  sup- 
posing the  velocity  of  sound  to  be  a  ft.  per  sec. 

Ans.     \    at  +  §- %    . 

LV  *g      V2gJ 

23.  A  stone  is  dropped  into  a  well,  and  after  3  sees,  the 
sound  of  the  splash  is  heard.  Find  the  depth  to  the 
surface  of  the  water,  the  velocity  of  sound  being  1127  ft. 
per  sec.  Ans.  132.9. 

24.  A  body  is  simultaneously  impressed  with  three 
uniform  velocities,  one  of  which  would  cause  it  to  move 
10  ft.  North  in  2  sees. ;  another  12  ft.  in  one  sec.  in  the 
same  direction ;  and  a  third  21  ft.  South  in  3  sees.  Where 
will  the  body  be  in  5  sees.  ?  Ans.  50  ft.  North. 

25.  A  boat  is  rowed  across  a  river  1J  miles  wide,  in  a 
direction  making  an  angle  of  87°  with  the  bank.  The 
boat  travels  at  the  rate  of  5  miles  an  hour,  and  the  river 
runs  at  the  rate  of  2.3  miles  an  hour.  Find  at  what  point 
of  the  opposite  bank  the  boat  will  land,  if  the  angle  of  87° 
be  made  against  the  stream. 

Ans.  898  yards  down  the  stream  from  the  opposite 
point. 

26.  A  body  moves  with  a  velocity  of  10  ft.  per  sec.  in  a 
given  direction  ;  find  the  velocity  in  a  direction  inclined  at 
an  angle  of  30°  to  the  original  direction. 

Ans.  5  VS  ft.  per  sec. 


EXAMPLES.  253 

27.  A  smooth  plane  is  inclined  at  an  angle  of  30°  to  the 
horizon  ;  a  body  is  started  up  the  plane  with  the  velocity 
5g;  find  when  it  is  distant  9g  from  the  starting  point. 

Ans.  2,  or  18  sees. 

28.  The  angle  of  a  plane  is  30° ;  find  the  velocity  with 
which  a  body  must  be  projected  up  it  to  reach  the  top, 
the  length  of  the  plane  being  20  ft. 

Ans.  8  a/10  ft.  per  sec. 

29.  A  body  is  projected  down  a  plane,  the  inclination  of 
which  is  45°,  with  a  velocity  of  10  ft.;  find  the  space 
described  in  2 J  sees.  Ans.  95.7  ft.  nearly. 

30.  A  steam-engine  starts  on  a  downward  incline  of 
1  in  200*  with  a  velocity  of  7J  miles  an  hour  neglecting 
friction  ;  find  the  space  traversed  in  two  minutes. 

Ans.  824  yards. 

31.  A  body  projected  up  an  incline  of  1  in  100  with  a 
velocity  of  15  miles  an  hour  just  reaches  the  summit ;  find 
the  time  occupied.  Ans.  68.75  sees. 

32.  From  a  point  in  an  inclined  plane  a  body  is  made  to 
slide  up  the  plane  with  a  velocity  of  16.1  ft.  per  sec.  (1) 
How  far  will  it  go  before  it  comes  to  rest,  the  inclination 
of  the  plane  to  the  horizon  being  30°  ?  (2)  Also  how  far 
will  the  body  be  from  the  starting  point  after  5  sees,  from 
the  beginning  of  motion  ? 

Ans.  (1)  8.05  ft.;  (2)  120.75  ft.  lower  down. 

33.  The  inclination  of  a  plane  is  3  vertical  to  4  hori- 
zontal ;  a  body  is  made  to  slide  up  the  incline  with  an 
initial  velocity  of  36  ft.  a  sec;  (1)  how  far  will  it  go  before 
beginning  to  return,  and  (2)  after  how  many  seconds  will 
it  return  to  its  starting  point  ? 

Ans.   (1)  33 J  ft.;  (2)  3 J  sees. 

*  An  incline  of  1  in  200  means  here  1  foot  vertically  to  a  length  of  200  ft.,  though 
it  is  used  by  Engineers  to  mean  1  foot  vertically  to  200  ft.  horizontally. 


254  EXAMPLES. 

34.  There  is  an  inclined  plane  of  5  vertical  to  12  hori- 
zontal, a  body  slides  down  52  ft.  of  its  length,  and  then 
passes  without  loss  of  velocity  on  to  the  horizontal  plane; 
after  how  long  from  the  beginning  of  the  motion  will  it  be 
at  a  distance  of  100  ft.  from  the  foot  of  the  incline? 

Ans.  5.7  sees. 

35.  A  body  is  projected  up  an  inclined  plane,  whose 
length  is  10  times  its  height,  with  a  velocity  of  30  ft.  per 
sec. ;  in  what  time  will  its  velocity  be  destroyed  ? 

Ans.  9|  sees.,  if  ^  —  32. 

36.  A  body  falls  from  rest  down  a  given  inclined  plane; 
compare  the  times  of  describing  the  first  and  last  halves 
of  it-  Ans.  As  1  :  V2  -  1. 

37.  Two  bodies,  projected  down  two  planes  inclined  to 
the  horizon  at  angles  of  45°  and  60°,  describe  in  the  same 
time  spaces  respectively  as  \/2  :  Vs  ;  find  the  ratio  of  the 
initial  velocities  of  the  projected  bodies. 

Ans.   V%  :  V3. 

38.  Through  what  chord  of  a  circle  must  a  body  fall  to 
acquire  half  the  velocity  gained  by  falling  through  the 
diameter  ? 

Ans.  The  chord  which  is  inclined  at  60°  to  the  vertical. 

39.  Find  the  velocity  with  which  a  body  should  be  pro- 
jected down  an  inclined  plane,  I,  so  that  the  time  of 
running  down  the  plane  shall  be  equal  to  the  time  of 
falling  down  the  height,  h. 

(I — h  sin  cA 

40.  Find  the  inclination  of  this  plane,  when  a  velocity 
of  f  ths  that  due  to  the  height  is  sufficient  to  render  the 
times  of  running  down  the  plane,  and  of  falling  down  the 
height,  equal  to  each  other-  Ans.  30°. 


EXAMPLES.  255 

41.  Through  what  chord  of  a  circle,  drawn  from  the 
bottom  of  the  vertical  diameter  must  a  body  descend,  so  as 

to  acquire  a  velocity  equal  to   -th   part   of   the  velocity 

acquired  in  falling  down  the  vertical  diameter  ? 

Ans.  If  6  denote  the  angle  between  the  required  chord 

and  the  vertical  diameter  cos  0  =  -  • 

n 

42.  Find  the  inclination,  6,  of  the  radius  of  a  circle  to 
the  vertical,  such  that  a  body  running  down  will  describe 
the  radius  in  the  same  time  that  another  body  requires  to 
fall  down  the  vertical  diameter.  Ans.  0  =  60°. 

43.  Find  the  inclination,  0,  to  the  vertical  of  the  diam- 
eter down  which  a  body  falling  will  describe  the  last  half 
in  the  same  time  as  the  vertical  diameter. 

3  V2  — 4 


Ans.  cos  6 


2V% 


44.  Show  that  the  times  of  descent  down  all  the  radii  of 
curvature  of  the  cycloid  (Fig.  40,  Calculus)  are  equal ;  that 

/8r" 
is,  the  time  down  PQ  is  equal  to  the  time  down  O'A  =  y  —  • 

45.  Find  the  inclination,  0,  to  the  horizon  of  an  inclined 
plane,  so  that  the  time  of  descent  of  a  particle  down  the 
length  may  be  n  times  that  down  the  height  of  the  plane. 

Ans.  0  =  sin-1-- 
n 

46.  Find  the  line  of  quickest  descent  from  the  focus  to 
a  parabola  whose  axis  is  vertical  and  vertex  upwards,  and 
show  that  its  length  is  equal  to  that  of  the  latus  rectum. 

47.  Find  the  line  of  quickest  descent  from  the  focus  of  a 
parabola  to  the  curve  when  the  axis  is  horizontal. 


25  G  EXAMPLES. 

48.  Find  geometrically  the  line  of  quickest  descent  (1) 
from  a  point  within  a  circle  to  the  circle  ;  (2)  from  a  circle 
to  a  point  without  it. 

49.  Find  geometrically  the  straight  line  of  longest 
descent  from  a  circle  to  a  point  without  it,  and  which 
lies  below  the  circle. 

50.  A  man  six  feet  high  walks  in  a  straight  line  at  the 
rate  of  four  miles  an  hour  away  from  a  street  lamp,  the 
height  of  wThich  is  10  feet;  supposing  the  man  to  start 
from  the  lamp-post,  find  the  rate  at  which  the  end  of  his 
shadow  travels,  and  also  the  rate  at  which  the  end  of  his 
shadow  separates  from  himself. 

Aiis.  Shadow  travels  10  miles  an  hour,  and  gains  on 
himself  6  miles  an  hour. 

51.  Two  bodies  fall  in  the  same  time  from  two  given 
points  in  space  in  the  same  vertical  down  two  straight 
lines  drawn  to  any  point  of  a  surface ;  show  that  the  sur- 
face is  an  equilateral  hyperboloid  of  revolution,  having  the 
given  points  as  vertices. 

52.  Find  the  form  of  a  curve  in  a  vertical  plane,  such 
that  if  heavy  particles  be  simultaneously  let  fall  from  each 
point  of  it  so  as  to  slide  freely  along  the  normal  at  that 
point,  they  may  all  reach  a  given  horizontal  straight  line  at 
the  same  instant. 

53.  Show  that  the  time  of  quickest  descent  down  a  focal 
chord  of  a  parabola  whose  axis  is  vertical  is 


where  I  is  the  latus  rectum. 


^ 


54.   Particles  slide  from  rest  at  the  highest  point  of  a 
vertical  circle  down  chords,  and  arc  then  allowed  to  move 


EXAMPLES.  257 

freely  ;  show  that  the  locus  of  the  foci  of  their  paths  is  a 
circle  of  half  the  radius,  and  that  all  the  paths  bisect  the 
vertical  radius. 

55.  If  the  particles  slide  down  chords  to  the  lowest  point, 
and  be  then  suffered  to  move  freely,  the  locus  of  the  foci  is 
a  cardioid. 

56.  Particles  fall  down  diameters  of  a  vertical  circle  ;  the 
locus  of  the  foci  of  their  subsequent  paths  is  the  circle. 


CHAPTER     II. 

CURVILINEAR     MOTION. 

147.  Remarks  on  Curvilinear  Motion. — The  mo- 
tion, which  was  considered  in  the  last  chapter,  was  that  of 
a  particle  describing  a  rectilinear  path.  In  this  chapter  the 
circumstances  of  motion  in  which  the  path  is  curvilinear 
will  be  considered.  The  conception  and  the  definition  of 
velocity  and  of  acceleration  which  were  given  in  Arts.  134, 
135,  are  evidently  as  applicable  to  a  particle  describing  a 
curvilinear  path  as  to  one  moving  along  a  straight  line ; 
and  consequently  the  formulae  for  velocity  in  Arts.  142, 143, 
are  applicable  either  to  rectilinear  or  to  curvilinear  motion. 
In  the  last  chapter  the  effects  of  the  composition  and  the 
resolution  of  velocities  were  considered,  when  the  path 
taken  by  the  particle  in  consequence  of  them  was  straight ; 
we  have  now  to  investigate  the  effects  of  velocities  and  of 
accelerations  in  a  more  general  way. 

148.  Composition  of  Uniform  Velocity  and  Ac- 
celeration.— Suppose  a  body  tends  to  move  in  one  direc- 
tion with  a  uniform  velocity  which  would  carry  it  from  A 
to  B  in  one  second,  and  also  subject  to  an 
acceleration  that  wo  aid  carry  it  from  A 
to  0  in  one  second ;  then  at  the  end  of 
the  second  the  body  will  be  at  D,  the 
opposite  end  of  the  diagonal  of  the  par-  \  Fig  77 
allelogram  ABDC,  just  as  if  it  had  moved 

from  A  to  B  and  then  from  B  to  D  in  the  second,  but  the 
body  will  move  in  the  curve  and  not  along  the  diagonal. 
For,  the  body  in  its  motion  is  making  progress  uniformly 
in  the  direction  AB,  at  the  same  rate  as  if  it  had  no  other 
motion ;  and  at  the  came  time  it  is  being  accelerated  in  the 


COMPOSITION  OF  ACCELERATIONS.  259 

direction  AC,  as  fast  as  if  it  had  no  other  motion.  Hence 
the  body  will  reach  D  as  far  from  the  line  AC  as  if  it  had 
moved  over  AB,  and  as  far  from  AB  as  if  it  had  moved 
over  AC;  but  since  the  velocity  along  AC  is  not  uniform, 
the  spaces  described  in  equal  intervals  of  times  will  not  be 
equal  along  AC  while  they  are  equal  along  AB,  and  there- 
fore the  points  a19  a2,  a3,  will  not  be  in  a  straight  line.  In 
this  case,  therefore,  the  path  is  a  curve. 

149.  Composition  and  Resolution  of  Accelera- 
tions.— If  a  body  is  subject  to  two  different  accelerations 
in  different  directions  the  sides  of  a  parallelogram  may  be 
taken  to  represent  the  Component  Accelerations,  and 
the  diagonal  will  represent  the  Resultant  Acceleration, 
although  the  path  of  the  body  may  be  along  some  other 
line. 

Rem. — These  results  with  those  of  Arts.  142,  143,  may  be 
summed  up  in  one  general  law :  When  a  body  tends  to 
move  with  several  different  velocities  in  different  directions, 
the  body  will  be,  at  the  end  of  any  given  time,  at  the  same 
point,  as  if  it  had  moved  with  each  velocity  separately. 
This  is  the  fundamental  law  of  the  composition  of  veloci- 
ties, and  it  shows  that  all  problems  which  involve  tenden- 
cies to  motion  in  different  directions  simultaneously,  may 
be  treated  as  if  those  tendencies  were  successive.* 

d?s 
If  —   be  the  acceleration  along  the  curve,  and  (x,  y,  z) 

be  the  place  of  the  moving  particle  at  the  time,  t,  it  is 

evident  that  the  component   accelerations  parallel   to  the 

cPx    dhj    d*z      _  ..         _ 

axes  are  -r^,  -j^ ,  -j-2-     Denoting  these  by  ax,  ccy,  az,  we 

have 

&x  _  <Py  _  cPz  _ 

dt>  ~  ax>    dP  ~  ay>    dP  "  "*; 


ind  Vccx2  +  ccy2  +  i(z2  is  the  resultant  acceleration. 


*  See  Remarks  on  Newton's  2d  law,  Art.  168. 


2G0  COMPOSITION  OF  ACCELERATIONS. 

Also  if  a,  ft  y,  be  the  angles  which  the  direction  oi 
motion  makes  with  the  axes,  we  have 

d2x       d?s 

_  =  __CoS«  =  ^. 

dry       d2s        _ 

-f  =  ^008  0  =  ,*; 


d2^       f?25 


The  acceleration  -=-j  is  not  generally  the  complete  resultant  of  the 

three  component  accelerations,  but  is  so  only  when  the  path  is  a 
straight  line  or  the  velocity  is  zero.     It  is,  however,  the  only  part 

of  their  resultant  which  has  any  effect  on  the  velocity.     —-^  is  the 

sum  of  the  resolved  parts  of  the  component  accelerations  in  the  direc- 
tion of  motion,  as  the  following  identical  equation  shows : 

dh  _  dx  drx      dy  d?y      dz  d?z 
dt*  ~  ds'dP  +  ds'dt*  +  ds' dt* 

which  follows  immediately  from  (1)  of  Art.  143    by  differentiation. 

Accelerations  are  therefore  subject  to  the  same  laws  of  composition 

and  resolution  as  velocities  ;  and  consequently  the  acceleration  of  the 

particle  along  any  line  is  the  sum  of  the  resolved  parts  of  the  axial 

d-s 
accelerations  along  that  line.   Thus  to  find  — ,  the  acceleration  along  s, 

~ 'lias  to  be  multiplied  by  j- ,  which  is  the  direction-cosine  of  the 

small  arc  ds.  The  other  part  of  the  resultant  is  at  right  angles  to 
this,  and  its  only  effect  is  to  change  the  direction  of  the  motion  of  the 
point.  (See  Tait  and  Steele's  Dynamics  of  a  Particle,  also  Thomson 
and  Tait's  Nat.  Phil.) 

The  following  are  examples  in  which  the  preceding  ex- 
press! on s  are  applied  to  cases  in  which  the  laws  of  velocity 
and  of  acceleration  are  given. 


EXAMPLES.  2G1 


EXAMPLES. 

1.  A  particle  moves  so  that  the  axial  components  of  its 
velocity  vary  as  the  corresponding  co-ordinates ;  it  is 
required  to  find,  the  equation  of  its  path ;  and.  the  accel- 
erations along  the  axes. 

Here  J  =  kx;    f  =  %; 

.-.    ^  =  d-E  =  Mt; 
x        y 

if  (a,  b)  is  the  initial  place  of  the  particle, 
•  •.    x  =  aeM ;    y  =  leu' 

.    x  _  y 

a       o 

is  the  equation  of  the  path. 
And  the  axial  accelerations  are 

d2x       7.        cPy       J9 

2.  A  wheel  rolls  along  a  straight  line  with  a  uniform 
velocity  ;  compare  the  velocity  of  a  given  point  in  the  cir- 
cumference with  that  of  the  centre  of  the  wheel. 

Let  the  line  along  which  the  wheel  rolls  be  the  axis  of  x, 
and  let  v  be  the  velocity  of  its  centre;  then  a  point  in  its 
circumference  describes  a  cycloid,  of  which,  the  origin 
being  taken  at  its  starting  point,  the  equation  is 

X  =  a  vers-1  ^  —  (2ay  —  y2y; 


262  EXAMPLES. 

dx  _  dy  ds 

yi"  (2a  -  y)*  ~~  (2a)*' 

But      v  =  -77  («  vers"1^)  = -t  •  -£; 

^    ds  _ds    dy  ___  fiy\±v. 

dt       dy    dt  ~~  \a]      ' 

which  is  the  velocity  of  the  point  in  the  circumference  oi 
the  wheel.  Thus  the  velocity  of  the  highest  point  of  the 
wheel  is  twice  as  great  as  that  of  the  centre,  while  the 
point  that  is  in  contact  with  the  straight  line  has  no 
velocity.     (See  Price's  Anal.  Mech's.,  Vol.  I,  p.  416.) 

dx  dn 

3.  If  -jr  =  ky,    -77  =  hx,  show  that  the  path  is  an  equi« 

lateral  hyperbola  and  that  the  axial  components  are 
d?x       79       d2y       79 

4.  A  particle  describes  an  ellipse  so  that  the  ^-component 
of  its  velocity  is  a  constant,  a  ;  find  the  ^/-component  of  its 
velocity  and  acceleration,  and  the  time  of  describing  the 
ellipse. 

Let  the  equation  of  the  ellipse  be 

x*      y2 

—    -4-  —   —   1  • 

a2^  b2~  L> 

and  let  (x,  y)  be  the  position  of  the  particle  at  the  time  t ; 

dx  ,     dy  b2x 

then  Tt  =  a;    and    £  =-^3 

dy  _  dy    dx  _        ab2    x 
dt       dx    dt  "  a2     y9 

which  is  the  ^-component  of  the  velocity. 


EXAMPLE. 

d?y 

dt*  ~ 

aft 

y* 

* 

= 

#«2 

> 

26a 


Also 


hence  the  acceleration  parallel  to  the  axis  of  y  varies 
inversely  as  the  cube  of  the  ordinate  of  the  ellipse,  and  acts 
towards  the  axis  of  x,  as  is  shown  by  the  negative  sign. 

The  time  of  passing  from  the  extremity  of  the  minor 
axis  to  that  of  the  major  axis  is  found  by  dividing  a  by  a, 
the  constant  velocity   parallel   to   the   axis   of  x,  giving 

- ,  and  the  time  of  describing  the  whole  ellipse  is  — 
a  a 

If  the  orbit  is  a  circle  b  =  a,  and  the  acceleration  par- 
ens 
allel  to  the  axis  of  y  is =-• 

yZ 

If  the  velocity  parallel  to  the  y-axis  is  constant  and  equal 
to  j3,  then 

dx  '  a2(3    y# 

di  ~~  "~  ~¥  '  x  5 

dt*  ""       W5 
and  the  periodic  time  =  -^* 

x2      ifi 
5.  A  particle  describes  the  hyperbola  —2  —  %  =  *  »  ^n(^ 

(1)  the  acceleration  parallel  to  the  axis  of  #  if  the  velocity 
parallel  to  the  axis  of  y  is  a  constant,  Q,  and  (2)  find  the 
acceleration  parallel  to  the  ?/-axis  if  the  velocity  paralW  to 
the  z-axis  is  a  constant  «. 
(1)  Here  we  have 


£  =  ">  and  £■=#•* 


264  EXAMPLES. 

dx  _  dx    dy  _  fid?   y 
•*'    dt  "  dy'dt  ~  ~W'x 

which  is  the  velocity  parallel  to  the  #-axis. 

•  dy         dx 

A1  dte       &a2   Xdt~~ydt 

Also 


dt*         &  x* 

(Pa* 

hence  the  acceleration  parallel  to  the  #-axis  varies  inversely 
as  the  cube  of  the  abscissa,  and  the  ic-component  of  the 
velocity  is  increasing. 


(2) 

Here 

we  have 

dx 
dt 

=  «; 

d 

• 
•  • 

dy 
dt  ~~ 

&y 

dP  ~~ 

aW    X 
a*'y; 

ay; 

hence  the  acceleration  parallel  to  the  y-axis  is  negative  and 
the  y-component  of  the  velocity  is  decreasing. 

6.  A  particle  describes  the  parabola,  x^  -f  y*  =  eft,  with 
a  constant  velocity,  c ;  find  the  accelerations  parallel  to  the 
axes  of  x  and  y. 

Here  we  have  -^  =  c ; 

dt 

and  *?  =  =£  =    *    . 

**      r      (*  +  y)* 


EXAMPLES.  265 

dx*  _df        x  c2x    # 

dt2  ~  dt2    x  -f  y  ~~  x  -f-  y f 

.  %2  _  ds2       y  <?y 

dt2  -df2'x~+y~x~+l/; 

differentiating  we  get 

d?x  _     c2(ay)* 
dt2  ~  %(x-\-yf% 

cPy  _      c2  (ax)* 
dP  ="  2  (x  +  y)2 

7.  A  particle  describes  a  parabola  with  such  a  varying 
velocity  that  its  projection  on  a  line  perpendicular  to  the 
axis  is  a  constant,  v.  Find  the  velocity  and  the  accelera- 
tion parallel  to  the  axis. 

Let  the  equation  of  the  parabola  be 

y2  =  2px; 

then  |  =  v, 

,  dx  _  dx    dy vy  # 

dt  ""  <%    dt  ~~  p  ' 

which  is  the  velocity  parallel  to  x 

d?x       v* 
Also  -=-  =  -, 

dt2       p 

which  shows  that  the  particle  is  moving  away  from  the 
tangent  to  the  curve  at  the  vertex  with  a  constant  accelera- 
tion. 

12 


266 


PROJECTILE  IN   VACUO. 


Hence  as  the  earth  acts  on  particles  near  its  surface  with 
a  constant  acceleration  in  vertical  lines,  if  a  particle  is 
projected  with  a  velocity,  v,  in  a  horizontal  line  it  will  move 
in  a  parabolic  path. 

150.  Motion  of  Projectiles  in  Vacuo. — If  a  particle 
be  projected  in  a  direction  oblique  to  the  horizon  it  is 
called  a  Projectile,  and  the  path  which  it  describes  is  called 
its  Trajectory.  The  case  which  we  shall  here  consider  is 
that  of  a  particle  moving  in  vacuo  under  the  action  of 
gravity ;  so  that  the  problem  is  that  of  the  motion  of  a 
projectile  in  vacuo ;  and  hence,  as  gravity  does  not  affect 
its  horizontal  velocity,  it  resolves  itself  into  the  purely 
kinematic  problem  of  a  particle  moving  so  that  its  hori- 
zontal acceleration  is  0  and  its  vertical  acceleration  is  the 
constant,  g,  (Art.  140). 

151.  The  Path  of  a 
Projectile  in  Vacuo  is  a 
Parabola. — Let  the  plane 
in  which  the  particle  is  pro- 
jected be  the  plane  of  xy; 
let  the  axis  of  x  be  horizon- 
tal and  the  axis  of  y  vertical 
and  positive  upwards,  the 
origin  being  at  the  point  of 
projection;    let   the   velocity 

of  projection  —  v,  and  let  the  line  of  projection  be  inclined 
at  an  angle  a  to  the  axis  of  x,  so  that  v  cos  a,  and  v  sin  « 
are  the  resolved  parts  of  the  velocity  of  projection  along  the 
axes  of  x  and  y.  It  is  evident  that  the  particle  will  con- 
tinue to  move  in  the  plane  of  xy,  as  it  is  projected  in  it, 
and  is  subject  to  no  force  which  would  tend  to  withdraw 
it  from  that  plane. 

Let  (x,  y)  be  the  place,  P,  of  the  particle  at  the  time  i ; 
then  the  equations  of  motion  are 


S           D 

E 

1 

P 

A 

\  x 

/ 

M          C 

B 

k 

Fig. 

78 

PROJECTILE   TN    VACUO.  267 

dt*  ~     '    dP  -       ^ 

fche  acceleration  being  negative  since  the  ^-component  of 

the  velocity  is  decreasing. 

The  first  and  second  integrals  of  these   equations   will 

then   be,   taking  the  limits   corresponding   to  t  =  t  and 

t  =  0, 

dx  dy 

—  =  v  cos  a;  -jT  =s  v  sin  «  —  ^;  (1) 

#  =  t;  cos  at ;   y  =  #  sin  at  —  j#tf2.  (2) 

Equations  (1)  and  (2)  give  the  coordinates  of  the  particle 
and  its  velocity  parallel  to  either  axis  at  any  time,  t. 
Eliminating  t  between  equations  (2)  we  obtain 

y  =  x  tan  a  —       gX  —  (3) 

9  2v2  cos2  a  v  ' 

which  is  the  equation  of  the  trajectory,  and  shows  that  the 
particle  will  move  in  a  parabola. 

152.  The  Parameter ;  the  Range  H ;  the  Greatest 
Height  H;  Height  of  the  Directrix. — Equation  (3)  of 
Art.  151  may  be  written 

0       2v2  sin  a  cos  «  2^  cos2  a 

x2 x  = \ —  y, 

9  9        ' 

/        v*  sin  a  cos  «\2  2v2  cos2  a  /         v2  sin2  «\   ,„x 

or.  ^ . — j  = __  (y  — _ j.  (1) 

By  comparing  this  with  the  equation  of  a  parabola 
referred  to  its  vertex  as  origin,  we  find  for 

the  abscissa  of  the  vertex  =      sm  «  cos  n  .         ^ 


268  PROJECTILE  IN   VACUO. 

the  ordinate  of  the  vertex  =  — ;  (3) 

the  parameter  (latus  rectum)  = •        (4) 

And  by  transferring  the  origin  to  the  vertex  (1)  becomes 

-  2v2  cos2  a  fK3k 

+  *= g — y  (5) 

which  is  the  equation  of  a  parabola  with  its  axis  vertical 
and  the  vertex  the  highest  point  of  the  curve. 

The  distance,  OB,  between  the  point  of  projection  and 
the  point  where  the  projectile  strikes  the  horizontal  plane 
is  called  the  Range  on  the  horizontal  plane,  and  is  the 
value  of  x  when  y  =  0.  Putting  y  =  0  in  (3)  of  A <t.  151 
and  solving  for  x,  we  get 

the  horizontal  range  E  =  OB  = ;  (6) 


which  is  evident,  also,  geometrically,  as  OB  =  200 ;  that 
is,  the  range  is  equal  to  twice  the  abscissa  of  the  vertex 

It  follows  from  (6)  that  the  range  is  the  greatest,  Tor  a 
given  velocity  of  projection,  when  a  =  45°,  in  which  case 

the  ran2:e  =  —  • 
9 

Also  it  appears  from  (6)  that  the  range  is  the  same  when 
a  is  replaced  by  its  complement ;  that  is,  for  the  same 
velocity  of  projection  the  range  is  the  same  for  two  differ- 
ent angles  that  are  complements  of  each  other.  If  a  =  45° 
the  two  angles  become  identical,  and  the  range  is  a 
maximum. 

OA  is  called  the  greatest  height,  H,  of  the  projectile,  and 

is  given  by  (3)  which,  when  a  =  45°  becomes  —  •  (7) 


VELOCITY  OF  THE  PROJECTILE.  269 

The  height  of  the  directrix  =  CD 

v2  sin2  a       4  2v2  cos2  a        v2 

=  -W-  +  i—J-  =  *s  (8) 

Hence  when  a  =  45°  the  focus  of  the  parabola  lies  in 
the  horizontal  line  through  the  point  of  projection. 

153.  The  Velocity  of  the  Particle  at  any  Point  of 
its  Path. — Let  V  be  the  velocity  at  any  point  of  its  path, 

then        V2  =  (^)V  (^)2,  or  by  (1)  of  Art.  151 

=  v2  cos2  a  -f  (v2  sin2  a  —  2v  sin  agt  +  g2t2) 
=  v2  —  2gy. 

To  acquire  this  velocity  in  falling  from  rest,  the  particle 

V2 
must  have  fallen  through  a  height  — ,  (6)  of  Art.  140,  or 

9 
its  equal 

^-y  =  M8-MP by  (8) 

=  PS. 

Hence,  the  velocity  at  any  point,  P,  on  the  curve  is  that 
which  the  particle  would  acquire  in  falling  freely  in  vacuo 
down  the  vertical  height  SP;  that  is,  in  falling  from  the 
directrix  to  the  curve ;  and  the  velocity  of  projection  at  0 
is  that  which  the  particle  would  acquire  in  falling  freely 
through  the  height  CD.  The  directrix  of  the  parabola  is 
therefore  determined  by  the  velocity  of  projection,  and  is  at 
a  vertical  distance  above  the  point  of  projection  equal  to 
that  down  which  a  particle  falling  would  have  the  velocity 
of  projection. 

154.  The   Time   of  Flight,  T,  along  a  Horizontal 
Plane. — Put  y  =  0  in  (3)  of  Art,  151,  and  solve  for  x,  the 


2 TO  TIME    OF  FLIGHT   OF  PROJECTILE. 

a     i  •  ,  ^        i  %v*  sin  «  cos  a      -r,    ,   , ,     , 

values  of  which  are  0  and But  the  horizon 

9 

,         -  •  T1  ,7        ,.  r     n-     i  i  %V  Sin    « 

tal  velocity  is  v  cos  «.     Hence  the  time  of  flight  = 

which  varies  as  the  sine  of  the  inclination  to  the  axis  of  x. 

155.  To  Find  the  Point  at  which  a  Projectile  will 
Strike  a  Given  Inclined  Plane  passing  through  the 
Point  of  Projection,  and  the  Time  of  Flight— Let  the 

inclined  plane  make  an  angle  j3  with  the  horizon  ;  it  is 
evident  that  we  have  only  to  eliminate  y  between  y  =  x  tan 
(3  and  (3)  of  Art,  151,  which  gives  for  the  abscissa  of  the 
point  where  the  projectile  meets  the  plane 


2v2  cos  a  sin  (a  —  (3)  # 
and  the  ordinate  is 


1  g  cos  (3 


(i) 


_  2v2  cos  a  tan  (3  sin  (a  —  (3) 
^x  g  cos  (3 


Hence  the  time  of  flight 

y_     *i      _  9»  sin  («  —  ft) 

v  cos  «  #  cos  j3  * 

156.  The  Direction  of  Projection  which  gives  the 
Greatest  Range  on  a  Given  Plane. — The  range  on  the 
horizontal  plane  is 

v2  sin  2cc 

which  for  a  given  value  of  v  is  greatest  when  a  =  -  (Art 

152). 
The  range  on  the  inclined  plane  =  xt  sec  (3 

_  2v%  cos  a  sin  («  —  (3)^  . 

#  cos2  j3  *  ' 


ANGLE   OF  ELEVATION   OF  PROJECTILE.  271 

To  find  the  value  of  a  which  makes  this  a  maximum,  we 
must  equate  to  zero  its  derivative  with  respect  to  «,  which 
gives 

cos  (2«  —  (3)  =  0; 

and  hence  «  —  j3  ==  ^J  U  —  0),  (3) 

which  is  the  angle  which  the  direction  of  projection  makes 
with  the  inclined  plane  when  the  range  is  a  maximum; 
that  is,  the  projection  bisects  the  angle  between  the 
inclined  plane  and  the  vertical. 

In  this  case  by  substituting  in  (1)  the  values  of  «  and 
(a  —  (3)  as  given  in  (2)  and  (3)  and  reducing,  we  get 

the  greatest  range  =  sin  ^  ■  (4) 

157.  The  angle  of  Elevation  so  that  the  Particle 
may  pass  through  a  Given  Point. — From  Art.  152, 
there  are  two  directions  in  which  a  particle  may  be  pro- 
jected so  as  to  reach  a  given  point ;  and  they  are  equally 

inclined  to  the  direction  of  projection  («  =  -)• 

Let  the  given  point  lie  in  the  plane  which  makes  an 
angle  j3  with  the  horizon,  and  suppose  its  abscissa  to  be  h  ; 
then  we  must  have  from  (1)  of  Art.  155 

cos  a  sin  (a  —  p)  =  h. 


g  cos  13 


If  «'  and  a"  be  the  two  values  of  a  which  satisfy  this 
equation,  we  must  have 

cos  «'  sin  («'  —  (3)  =  cos  a"  sin  (a"  —  (3) ; 


OiVO 


EQUATION  OF  TRAJECTORY,   SECOND  METHOD. 


and  therefore  a"  —  (3  =  -  —  «', 

or  rfV—lg  +  pJ-tg  +  nJ-W.  (1) 

But  each  member  of  (1)  is  the  angle  between  one  of  the 
directions  of  projection  and  the  direction  for  the  greatest 
range  [Art.  156,  (2)].  Hence,  as  in  Art.  152,  the  two 
directions  of  projection  which  enable  the  particle  to  pass 
through  a  point  in  a  given  plane  through  the  point  of  pro- 
jection, are  equally  inclined  to  the  direction  of  projection 
for  the  greatest  range  along  that  plane.  (See  Tait  and 
Steele's  Dynamics  of  a  Particle,  p.  89.) 

158.  Second  Method  of  Finding  the  Equation  of 
the  Trajectory. — By  a  somewhat  simpler  method  than 
that  of  Art.  151,  we  may  find  the  equation  of  the  path  of 
the  projectile  as  the  resultant  of  a  uniform  velocity  and  an 
acceleration  (Art.  148). 

Take  the  direction  of  projection  (Fig.  78)  as  the  axis  of 
x,  and  the  vertical  downwards  from  the  point  of  projection 
as  the  axis  of  y.  Then  (Art.  149,  Eem.)  the  velocity,  v, 
due  to  the  projection,  will  carry  the  particle,  with  uniform 
motion,  parallel  to  the  axis  of  x,  while  at  the  same  time,  it 
is  carried  with  constant  acceleration,  g,  parallel  to  the  axis 
of  y.  Hence  at  any  time,  t,  the  equations  of  motion  along 
the  axes  of  a?  and  y  respectively  are 

x  =  vt, 

y  =  fe^- 

That  is,  if  the  particle  were  moving  with  the  velocity  v, 
alone,  it  would  in  the  time  t,  arrive  at  Q ;  and  if  it  were 
then  to  move  with  the  vertical  acceleration  g  alone  it  would 
in  the  same  time  arrive  at  P ;  therefore  if  the  velocity  v 


EXAMPLES.  273 

and  the  acceleration  g  are  simultaneous,  the  particle  will  ir 
the  time  /  arrive  at  P  (Art.  149,  Rem). 
Eliminating  t  we  have 

which  is  the  equation  of  a  parabola  referred  to  a  diameter 
and  the  tangent  at  its  vertex.  The  distance  of  the  origin 
from   the   directrix,   being   Jth  of   the  coefficient  of  y,  is 

— ,  as  in  Art.  152,  (8). 

EXAMPLES. 

1.  From  the  top  of  a  tower  two  particles  are  projected  at 
angles  a  and  (3  to  the  horizon  with  the  same  velocity,  v,  and 
both  strike  the  horizontal  plane  passing  through  the  bot- 
tom of  the  tower  at  the  same  point;  find  the  height  of 
the  tower. 

Let  h  =  the  height  of  the  tower;  v  —  the  velocity  of 
projection ;  then  if  the  particles  are  projected  from  the 
edge  of  the  top  of  the  tower,  and  x  is  the  distance  from  the 
bottom  of  the  tower  to  the  point  where  they  strike  the 
horizontal  plane  we  have  from  (3)  of  Art.  151 

—  li  =  x  tan  a  —  |^  (1  +  tan2  «),  (1) 

-  h  =  x  tan  j3  -  |^  (1  +  tan2  /3),  (2) 

by  subtraction 

2v2  2v2  cos  a  cos  13 

g  (tan  a  -f  tan  (3)        g  sin  (ce  -f-  (3)  y 

which  in  (1)  or  (2)  gives 

,  __  2v2  cos  a  cos  (3  cos  (a  +  (3) 
g  [sin  («  +  ^)P~ 


274  VELOCITY  OF  DISCHARGE    OF  SHELLS. 

2.  Particles  are  projected  with  a  given  velocity  in  all 
lines  in  a  vertical  plane  from  the  point  0;  it  is  required  to 
find  the  locus  of  their  highest  points. 

Let  (x,  y)  be  the  highest  point ;  then  from  (2)  and  (3)  of 
Art.  152,  we  have 

v2  sin  a  cos  a 
x  =  ; 

9 
v2  sin2  a 

therefore      sin2  a  =  -—-,  and  cos2  a  =  ~-^— 
v2  2v2y 

Adding  4f  +  x2  =  ^-; 

which  is  the  equation  of  an  ellipse,  whose  major  axis  ~  — ; 

v2 
and  the  minor  axis  =  ^-;  and  the  origin  is  at  the  extremity 

of  the  minor  axis. 

3.  Find  the  angle  of  projection,  <*,  so  that  the  area  con- 
tained between  the  path  of  the  projectile  and  the  hori- 
zontal line  may  be  a  maximum,  and  find  the  value  of  the 
maximum  area. 

Ans.  a  =  60°  and  Max.  Area  =  -^  (3)^. 

4.  Find  the  ratio  of  the  areas  A1  and  A2  of  the  two 

parabolas  described  by  projectiles  whose  horizontal  ranges 

are  the  same,  and  the  angles   of  projection  are  therefore 

complements  of  each  other.  At     _  ,     2 

Ji'iiSt  ~i —  —  tan  cc, 

159.  Velocity  of  Discharge  of  Balls  and  Shells 
from  the  Mouth  of  a  Gun. — As  the  result  of  numerous 


ANGULAR    VELOCITY.  275 

experiments  made  at  Woolwich,  the  following  formula  was 
regarded  as  a  correct  expression  for  the  velocity  of  balls  and 
shells,  on  quitting  the  gun,  and  fired  with  moderate 
charges  of  powder,  from  the  pieces  of  ordnance  commonly 
used  for  military  purposes : 


v  =  1600 


/3P 


where  v  is  the  velocity  in  feet  per  second,  P  the  weight  of 
the  charge  of  powder,  and  W  the  weight  of  the  ball. 

For  the  investigation  of  the  path  of  a  projectile  in  the 
atmosphere,  see  Chap.  I  of  Kinetics. 

160.  Angular  Velocity,  and  Angular  Accelera- 
tion.— Hitherto  the  method  of  resolving  velocities  and 
accelerations  along  two  rectangular  axes  has  been  employed. 
It  remains  for  us  to  investigate  the  kinematics  of  a  particle 
describing  a  curvilinear  path,  from  another  point  of  view 
and  in  relation  to  another  system  of  reference.  Before  wre 
consider  velocities  and  accelerations  in  reference  to  a 
system  of  polar  co-ordinates,  it  is  necessary  to  enquire  into 
a  mode  of  measuring  the  angular  velocity  of  a  particle. 

Angular  Velocity  may  he  defined  as  the  rate  of  angular 
motion.  Thus  let  (r,  6)  be  the  position  of  the  point  P,  and 
suppose  that  the  radius  vector  has  revolved  uniformly 
through  the  angle  6  in  the  time  t,  then  denoting  the 
angular  velocity  by  g>,  we  shall  have,  as  in  linear  velocity 
(Art.  7) 

6 

"  =  r 

If  however  the  radius  vector  does  not 
revolve  uniformly  through  the  angle  6 
we  may  always  regard  it  as  revolving 
uniformly  through  the  angle  dO  in  the 
infinitesimal  of  time  dt  ;  hence  we  shall 
have  as  the  proper  value  of  w, 


2'. 

Fig.79 


276  EXAMPLES. 

dO 

w  =  di-  (i> 

Hence,  whether  the  angular  Telocity  be  uniform  or 
variable,  it  is  the  ratio  of  the  angle  described  by  the  radio  a 
vector  in  a  given  time  to  the  time  in  which  it  is  described; 
thus  the  increase  of  the  angle,  in  angular  velocity,  takei 
the  place  of  the  increase  of  the  distance  from  a  fixed  point, 
in  linear  velocity,  (Art.  7). 

Angular  Acceleration  is  the  rate  of  increase  of  angular 
velocity ;  it  is  a  velocity  increment,  and  is  measured  in  the 
same  way  as  linear  acceleration  (Art.  9).  Thus,  whether 
the  angular  acceleration  is  uniform  or  variable,  it  may 
always  be  regarded  as  uniform  during  the  infinitesimal  of 
time  dt  in  which  time  the  increment  of  the  velocity  will  be 
d(*).  Hence  denoting  the  angular  acceleration  at  any  time, 
t,  by  0,  we  have 


do d 

dt   ~  di 


and  thus,  whether  the  increase  of  angular  velocity  is 
uniform  or  variable,  the  angular  acceleration  is  the  increase 
of  angular  velocity  in  a  unit  of  time. 

The  following  examples  are  illustrations  of  the  preceding 
mode  of  estimating  velocities  and  accelerations. 

EXAMPLES. 

1.  If  a  particle  is  placed  on  the  revolving  line  at  the 
distance  r  from  the  origin,  and  the  line  revolves  with  a 
uniform  angular  velocity,  w,  the  relation  between  the  linear 
velocity  of  the  particle  and  the  angular  \elocity  may  thus 
be  found.  . 


EXAMPLES.  277 

Let  dd  be  the  angle  through  which  the  radius  revolves  in 
the  time  dt,  and  let  ds  be  the  path  described  by  the  particle, 
so  that  ds  =  rdd ; 

ds  dd 

theU  di  =  rdt=0)r> 

so  that  the  linear  velocity  varies  as  the  angular  velocity  and 
the  length  of  the  radius  jointly. 

2.  If  the  angular  acceleration  is  a  constant,  as  0 ;  then 
from  (2)  we  have 

d*d 
df*=<t>> 

...     _  =  #  +  «„ 

and  .•.    6  =  %<pt2  +  u0t  +  0O, 

where  w0  and  60  are  the  initial  values  of  w  and  6. 

Hence  if  a  line  revolves  from  rest  with  a  constant  angular 
acceleration,  we  have 

and  the  angle  described  by  it  varies  as  the  square  of  the 
time. 

3.  If  a  particle  revolves  in  a  circle  uniformly,  and  its 
place  is  continually  projected  on  a  given  diameter,  the 
linear  acceleration  along  that  diameter  varies  directly  as 
the  distance  of  the  projected  place  from  the  centre. 

Let  to  be  the  constant  angular  velocity,  0  the  angle 
between  the  fixed  diameter  and  the  radius  drawn  from  the 
centre  to  its  place  at  the  time  t,  x  the  distance  of  this 
projected  place  from  the  centre.  Then,  calling  a  the 
radius  of  the  circle,  we  have 

x  =  a  cos  0, 


278  EXAMPLES. 

dx  .    add 

TT  —  —  a  sin  u  -=-  =  —  «w  sin  0 ; 

at  dt  ' 

d2x  dd 

which  proves  the  theorem. 

4.  If  the  angular  acceleration  varies  as  the  angle 
generated  from  a  given  fixed  line,  and  is  negative,  find  the 
angle. 

Here  the  equation  which  expresses  the  motion  is  of  the 
form 

%  =  -m. 

dt2 
Calling  a  the  initial  value  of  0  we  find  for  the  result 
6  =  a  cos  kt. 

5.  If  a  particle  revolves  in  a  circle  with  a  uniform 
velocity,  show  that  its  angular  velocity  about  any  point  in 
the  circumference  is  also  uniform,  and  equal  to  one-half  of 
what  it  is  about  the  centre. 

At  present  this  is  sufficient  for  the  general  explanation 
of  angular  velocity  and  angular  acceleration.  We  shall 
return  to  the  subject  in  Chap.  7,  Part  III.,  when  we  treat 
of  the  motion  of  rigid  bodies. 

161.  The  Component  Accelerations,  at  any  instant, 
Along,  and  Perpendicular  to  the  Radius  Vector. — 

Let  (r,  6)  (Fig.  79)  be  the  place  of  the  moving  particle,  P, 
at  the  time  t,  (x,  y)  being  its  place  referred  to  a  system  of 
rectangular  axes  having  the  same  origin,  and  the  z-axis 
coincident  with  the  initial  line.     Then 

x  =  r  cos  0  j  y  =:  r  sin  6 ;  (1) 


RADICAL  AND   TRANSVERSAL   ACCELERATIONS.  27i* 

*i       e                  dx       dr                     .     ndO  ' 

therefore            -37  =  -n  cos  6  —  r  sin  0  —  ;  (2) 

dt        dt                           dt  '  v  ' 


and 
dt* 


Similarly 
/ 


d*y       r#r         /^\2-|  .    .  ,  r^dr   dd  ,      ^01       afA. 


which  are  the  accelerations  parallel  to  the  axes  of  v  and  y. 
Eesolving  these  along  the  radius  vector  by  multiplying  (3) 
and  (4)  by  cos  0  and  sin  0  respectively,  since  accelerations 
may  be  resolved  and  compounded  along  any  line  the  same 
as  velocities  (Art.  149),  and  adding,  we  have 


dt* 


a      cVy   .     .       d?r         (dB\*  tKS 


which  is  the  acceleration  along  the  radius  vector.* 

Multiplying  (3)  and  (4)  by  sin  0  and  cos  0  respectively, 
and  subtracting  the  former  from  the  latter,  we  get 

d?y        ,       cPx   .    „       a  dr    dd         d2d 
JCOs0-WSm6  =  2di'dt+rdi> 

-ll(r2dl).  (6) 

-rdtX   dth  W 

which  is  the  acceleration  perpendicular  to  the  radius  vector. \ 

162.  The    Component   Accelerations,  at  any  in- 
stant, Along,  and  Perpendicular  to  the  Tangent. — 

I^t  (x>  y)  (Fig.  79)  be  the  place  of  the  moving  particle,  P, 
at  the  time  t,  and  s  the  length  of  the  arc  described  during 

*  Sometimes  called  the  radial  acceleration. 
t  Sometimes  called  the  transversal  acceleration. 


280  TANGENTIAL   ACCELERATION. 

that  time.    Then  the  accelerations  along  the  axes  of  x  and  3 

ctsX  d  u  dx  (In 

are  -rs  and  ~f  :  and  the  direction  cosines*  are  -=-  and  -f- • 

To  find  the  acceleration  along  the  tangent  we  must  multi- 
ply these  axial  accelerations  by  -j-  and  -—-,  respectively,  and 
add.     Thus  the  tangential  acceleration,  T,  is 

T  _  ^    dx      dhj    dy 

~  dt2  '  ds  "*"  dfi  '  ds'  K  } 

Since  ds2  =  dx2  -f  dy2,  therefore,  by  differentiation  we 
have 

ds  d?s  =  dx  d2x  4-  dy  d2y ; 

and  dividing  by  ds  dt2  we  get 


d2s  _  d2x    dx  |   d?y    dy 
dT 

which  in  (1)  gives 


dt2  ~  dt2  '  ds  +  dt2  '  ds1 


for  ^0  acceleration  along  the  tangent. 

Similarly  we  have  for  the  normal  acceleration,  N, 

^j. d2y    dx       d2x    dy 

~~  dt2     ds       dt2     ds 

(d?y  dx  —  &x  dy)    d& 
~  r/s3  '  dt2 

1    $s2 
= -rr2,  (by  Ex.  4,  p.  144,  Calculus), 

where  p  is  the  radius  of  curvature  ; 

*  Cosines  of  the  angles  which  the  tangent  makes  with  the  axes  of  x  and  y. 


NORMAL   A  Ci  'EL  Eli  A  T10N.  £81 

V2 

■■■    N  =  J,  (3) 

il  r  is  the  velocity  of  the  particle  at  the  point  (x,  y). 

Hence  at  any  point,  P,  of  the  trajectory,  if  tlie  accelera- 
tion is  resolved  along  the  tangent  to  the  curve  at  P  and 
along  the  normal,  the  accelerations  along  the  two  lines  are 
respectively 

d2s         ,     v* 

^2     and     — 
dt2  p 

163.  When  the  Acceleration  Perpendicular  to 
the  Radius  Vector  is  zero. — Then  from  (6)  of  Art.  161 
we  have 

r2  -j.  =  constant  =  h  suppose ; 


•"'    dt  ~  r2' 

dr 
di 

dr    dd       li     dr  # 

"  dd  '  di  ~  ? 8 '  dd ; 

dt2 

h2    d2r         h2  /dr\\ 
~~  i*  '  dd2  ~  ~  r»  \dd)  ; 

and 


which  in  (5)  of  Art.  161  gives 

the  acceleration  along  the  radius  vector 

—  ri(ld2       ~r*\dd)      r3'  '  ' 

an  expression  which  is  independent  of  t. 

This  may  be  put  into  a  more  convenient  form  as  follows: 

Let  r  =  -;  then 
u 

dr  1     du # 

dd  ~~  "~  w2 '  50  ' 


282  CONSTANT  ANGULAR    VELOCITY. 

'*'     dm"       u*'d&  +  w8W; 
which  in  (1)  and  reducing,  gives 
the  acceleration  along  the  radius  v<*».f«r 

=  _AV(g  +  «).  (2) 

From  these  two  formulae  the  law  of  acceleration  along 
the  radius  vector  may  be  deduced  when  the  curve  is  given, 
and  the  curve  may  be  deduced  when  the  law  of  accelera- 
tion along  the  radius  vector  is  given.  Examples  of  these 
processes  will  be  given  in  Chap.  (2),  Part  III. 

164.  When  the  Angular  Velocity  is  Constant- 
Let  the  angular  velocity  be  constant  =  w  suppose.    Then 

dS 

therefore  from  (5)  of  Art.  161 

the  acceleration  along  the  radius  vector 

=  £-«*  a) 

The  acceleration  perpendicular  to  the  radius  vector 

and  both  of  these  are  independent  of  0. 

The  following  example  is  an  illustration  of  these 
formulae  : 

A  particle  describes  a  path  with  a  constant  angular 
velocity,  and  without  acceleration  along  the  radius  vector ; 
find  (1)  the  equation  of  the  path,  and  (2)  the  acceleration 
perpendicular  to  the  radius  vector. 


CONSTANT  ANGULAR    VELOCITY.  283 

(1)  From    (1)    we   have,    from    the    conditions    of    the 
question, 

cPr 


dt>    w2r  =  °- 


Integrating  we  have 

civ 
if  r  =  a  when  -y-  =  0. 
at 

civ 
Therefore r  =  wdt\ 

if  r  =  a  when  t  =  0, 

.•.    r  =  |(H  +  r^).  (3) 

Also,  as  -^r  =  w,  therefore  0  =  ut,  ii  6  =  0  when  #  =  0. 
Substituting  this  value  of  ut,  we  have, 

r  =  |(«» +  «-»);  (4) 

which  is  the  path  described  by  the  particle. 

(2)  Let  Q  be  the  required  acceleration  perpendicular  to 
the  radius  vector,  then  from  (2)  we  have 

=  aw2  O'  —  e~w'),  from  (3) 


284  EXAMPLES. 

—  ad*  (ee  —  e~9) 

=  2w2(r2_  a2)i.  (5) 

which  is  the  acceleration  perpendicular  to  the  radius 
vector. 

The  preceding  discussion  of  Kinematics  is  sufficient  for 
this  work.  There  are  various  other  problems  which  might 
be  studied  as  Kinematic  questions,  and  inserted  here  ;  but 
we  prefer  to  treat  fchem  from  a  Kinetic  point  of  view. 

For  the  investigation  of  the  kinematics  of  a  particle 
describing  a  curvilinear  path  in  space,  see  Price's  Anal. 
Mech's,  Vol.  I,  p.  430,  also  Tait  and  Steele's  Dynamics  of 
a  Particle,  p.  12. 

EXAMPLES. 

1.  A  particle  describes  the  hyperbola,  xy  =  k2 ;  find  (1) 
the  acceleration  parallel  to  the  axis  of  x  if  the  velocity 
parallel  to  the  axis  of  y  is  a  constant,  j3,  and  (2)  find  the 
acceleration  parallel  to  the  axis  of  y  if  the  velocity  parallel 
to  the  axis  of  #  is  a  constant,  a. 

Am.   (1)-^^;  (2)  ^y\ 

2.  A  particle   describes   the   parabola,   y2  =  4ax ;    find 

the  acceleration  parallel  to  the  axis  of  y  if  the   velocity 

parallel  to  the  axis  of  x  is  a  constant,  «.        .              4a2a2 
1  Am. v-  * 

3.  A  particle  describes  the  logarithmic  curve,  y  =  ax; 
find  (1)  the  ^-component  of  the  acceleration  if  the  y-com- 
ponent  of  the  velocity  is  a  constant,  (3,  and  (2)  find  the 
^-component  of  the  acceleration  if  the  ^-component  of  the 
velocity  is  a  constant,  «. 

Am-  (a>-^^5(2)«»(iog«)«y. 


5.  A 


EXAMPLES.  285 

4.  A  particle  describes  the  cycloid,  the  starting  point 
being  the  origin;  find  (1)  the  ^-component  of  the  accel- 
eration it"  the  //-component  of  the  velocity  is  j3,  and  (2)  find 
the  //-component  of  the  acceleration  if  the  ^-component  of 

the  velocity  is  «.  Pay        t  aa? 

Ans.   {L)  —  ,  (4)  —-»' 

(2ay  -f)  i  V 

a  I  -        -  -\ 
particle  describes  a  catenary,  y  =  ~  ( ea  -f  e    a  I ; 

find  (1)  the  .^-component  of  the  acceleration  if  the  y-com- 
ponent  of  the  velocity  is  0,  and  (2)  find  the  ^-component 
of  the  acceleration  if  the  ^-component  of  the  velocity  is  a. 

6.  Determine  how  long  a  particle  takes  in  moving  from 

the  point  of  projection  to  the   further  end  of  the  latus 

rectum.  .        v  ,  . 

Ans.  -  (sin  «  +  cos  «). 
9 

7.  A  gun  was  fired  at  an  elevation  of  50°;  the  ball 
struck  the  ground  at  the  distance  of  2449  ft.;  find  (1)  the 
velocity  with  which  it  left  the  gun  and  (2)  the  time  of 
flight,     (g  =  32i). 

Ans.  (1)  282.8  ft.  per  sec;  (2)  13.47  sees. 

8.  A  ball  fired  with  velocity  u  at  an  inclination  a  to  the 

horizon,  just  clears  a  vertical  wall  which  subtends  an  angle, 

j3,  at  the  point   of  projection;   determine   the   instant  at 

which  the  ball  just  clears  the  wall. 

.        u  sin  a  —  Xqt  n 

Ans. —  =  tan  0. 

u  cos  a 

9.  In  the  preceding  example  determine   the  horizontal 

distance  between  the  foot  of  the  wall  and  the  point  where 

the  ball  strikes  the  ground.  .        2u2 

Ans.  —  cos2  a  tan  13. 
9 


286  EXAMPLES. 

10.  At  the  distance  of  a  quarter  of  a  mile  from  the  bot- 
tom of  a  cliff,  which  is  120  ft.  high,  a  shot  is  to  be  fired 
which  shall  just  clear  the  cliff,  and  pass  over  it  horizon- 
tally ;  find  the  angle,  cc,  and  velocity  of  projection,  v. 

Ans.  cc  =  10°  18';  v  =  490  ft.  per  sec. 

11.  When  the  angle  of  elevation  is  40°  the  range  is 
2449  ft. ;  find  the  range  when  the  elevation  is  29£°. 

Ans.  2131.5  ft. 

12.  A  body  is  projected  horizontally  with  a  velocity  of 
4  ft.  per  sec. ;  find  the  latus  rectum  of  the  parabola  de- 
scribed, (g  =  32).  Ans.   1  foot. 

13.  A  body  projected  from  the  top  of  a  tower  at  an  angle 
of  45°  above  the  horizontal  direction,  fell  in  5  sees,  at  a 
distance  from  the  bottom  of  the  tower  equal  to  its  altitude ; 
find  the  altitude  in  feet,  (g  =  32).  Ans.   200  feet. 

14.  A  ball  is  fired  up  a  hill  whose  inclination  is  15°; 
the  inclination  of  the  piece  is  45°,  and  the  velocity  of  pro- 
jection is  500  ft.  per  sec;  find  the  time  of  flight  before 
it  strikes  the  hill,  and  the  distance  of  the  place  where  it 
falls  from  the  point  of  projection.* 

Ans.  T  =  16.17  sees.;  E  =  1.121  miles. 

15.  On  a  descending  plane  w7hose  inclination  is  12°,  a 
ball  fired  from  the  top  hits  the  plane  at  a  distance  of  two 
miles  and  a  half,  the  elevation  of  the  piece  is  42°  ;  find  the 
velocity  of  projection.  Ans.  v  =  579.74  ft.  per  sec. 

16.  A  body  is  projected  at  an  iuclination  «  to  the  hori- 
zon; determine  when  the  motion  is  perpendicular  to  a 
plane  which  is  inclined  at  an  angle  /3  to  the  horizon. 

.       u  sin  a  —  at         ,       .' 

Ans. =  4-  cot  0. 

u  cos  a 

*  The  range  on  the  inclined  plane. 


EXAMPLES.  287 

17.  Calculate  the  maximum  range,  and  time  of  flight, 
on  a  descending  plane,  the  angle  of  depression  of  which  is 
15°,  the  velocity  of  projection  being  1000  ft.  per  sec. 

Ans.  Max.  range  =  7.98  miles  ;  T  =  51.34  sec. 

18.  With  what  velocity  does  the  ball  strike  the  plane  in 
the  lust  example  ?  Ans.   V  =  1303  feet. 

19.  If  a  ship  is  moving  horizontally  with  a  velocity 
=  dg,  and  a  body  is  let  fall  from  the  top  of  the  mast,  find 
its  velocity,  V,  and  direction,  0,  after  4  sees. 

Ans.  V  =  5g;  6  =  tan"1 §. 

20.  A  body  is  projected  horizontally  from  the  top  of  a 
tower,  with  the  velocity  gained  in  falling  down  a  space 
equal  to  the  height  of  the  tower ;  at  what  distance  from 
the  base  of  the  tower  will  it  strike  the  ground  ? 

Ans.   R  =  twice  the  height  of  the  tower. 

21.  Find  the  velocity  and  time  of  flight  of  a  body  pro- 
jected from  one  extremity  of  the  base  of  an  equilateral 
triangle,  and  in  the  direction  of  the  side  adjacent  to  that 
extremity,  to  pass  through  the  other  extremity  of  the  base. 

22.  (riven  the  velocity  of  sound,  V;  find  the  horizontal 
range,  when  a  ball,  at  a  given  angle  of  elevation,  «,  is  so 
projected  towards  a  person  that  the  ball  and  sound  of  the 
discharge  reach  him  at  the  same  instant. 

2  V2 

Ans, tan  «. 

9 

23.  A  body  is  projected  horizontally  with  a  velocity  of 
ig  from  a  point  whose  height  above  the  ground  is  16g ;  find 
the  direction  of  motion,  6,  (1)  when  it  has  fallen  half-way 
to  the  ground,  and  (2)  when  half  the  whole  time  of  falling 
has  elapsed.  ^  g  =  ^ .   (     $  =  ^    1 


288  EXAMPLES. 

24.  Particles  arc  projected  with  a  given  velocity,  v,  in 
all  lines  in  a  vertical  plane  from  the  point  0  ;  find  the  locus 
of  them  at  a  given  time,  t. 

Ans.  x2  +  (if  -j-  %gt?)2  =  v2l2,  which  is  the  equation  of  a 
circle  whose  radius  is  vt  and  whose  centre  is  on  the  axis  of 
y  at  a  distance  \gt%  below  the  origin. 

25.  How  much  powder  will  throw  a  13-inch  shell* 
4000  ft.  on  an  inclined  plane  whose  angle  of  elevation  is 
10°  40';  the  elevation  of  the  mortar  being  35°. 

Ans.  Charge  =  4.67  lbs. 

26.  A  projectile  is  discharged  in  a  horizontal  direction, 
with  a  velocity  of  450  ft.  per  sec,  from  the  summit  of  a 
conical  hill,  the  vertical  angle  of  which  is  120° ;  at  what 
distance  down  the  hillside  will  the  projectile  fall,  and  what 
will  be  the  time  of  flight? 

Ans.  Distance  =  2812.5  yards;  Time  =  16.23  sees. 

27.  A  guu  is  placed  at  a  distance  of  500  ft.  from  the  base 
of  a  cliff  which  is  200  ft.  high  ;  on  the  edge  of  the  cliff 
there  is  built  the  wall  of  a  castle  60  ft.  high ;  find  the 
elevation,  a,  of  the  gun,  and  the  velocity  of  discharge,  v, 
in  order  that  the  ball  may  graze  the  top  of  the  castle  wall, 
and  fall  120  ft.  inside  of  it. 

Ans.  a  =  53°  19' ;  v  =  165  ft.  per  sec. 

28.  A  piece  of  ordnance  burst  when  50  yards  from  a 
wall  14  ft.  high,  and  a  fragment  of  it,  originally  in  con- 
tact with  the  ground,  after  grazing  the  wall,  fell  6  ft. 
beyond  it  on  the  opposite  side ;  find  how  high  it  rose  in 
the  air.  Ans.  94  ft. 

*  The  weight  of  a  13-inch  shell  is  196  lbs. 


/= 


PART    III. 

KINETICS   (MOTION   AND   FORCE) 


CHAPTER    I. 

LAWS  OF  MOTION— MOTION  UNDER  THE  ACTION  OF 
A  VARIABLE  FORCE  — MOTION  IN  A  RESISTING 
MEDIUM. 

165.  Definitions. — Kinetics,  is  that  branch  of  Dynamics 
which  treats  of  the  motion  of  bodies  under  the  action  of 
forces. 

In  Part  I,  forces  were  considered  with  reference  to  the 
pressures  which  they  produced  upon  bodies  at  rest  (Art. 
15),  i.  e.,  bodies  under  the  action  of  two  or  more  forces 
in  equilibrium  (Art.  26).  In  Part  II  we  considered  the 
purely  geometric  properties  of  the  motion  of  a  point  or 
particle  without  any  reference  to  the  causes  producing  it, 
or  the  properties  of  the  thing  moved.  We  are  now  to 
consider  motion  with  reference  to  the  causes  which  produce 
it,  and  the  things  in  which  it  is  produced. 

The  student  must  here  review  Chapter  I,  Part  I,  and  obtain  clear 
conceptions  of  Momentum,  Acceleration  of  Momentum,  and  the  Kinetic 
measure  of  Force  (Arts.  12, 13, 19,  and  20),  as  this  is  necessary  to  a  full 
understanding  of  tbe  fundamental  laws  of  motion,  on  the  truth  of 
which  all  our  succeeding  investigations  are  founded. 

166.  Newton's  Laws  of  Motion.— The  fundamental 

13 


290  NEWTON* S  LAWS    OF  MOTION. 

principles  in  accordance  with  which  motion  takes  place  are 
embodied  in  three  statements,  generally  known  as  Newton-s 
Laivs  of  Motion.  These  laws  must  be  considered  as  resting 
on  convictions  drawn  from  observation  and  experiment, 
and  not  on  intuitive  perception.*  The  laws  are  the  fol- 
lowing : 

Law  I. — Every  body  continues  in  its  state  of  rest 
or  of  uniform  motion  in  a  straight  line,  except  in 
so  far  as  it  is  compelled  by  force  to  change  that 
state. 

Law  II. — Change  of  motion  is  proportional  to  the 
force  applied,  and  tahes  place  in  the  direction  of 
the  straight  line  in  which  the  force  acts. 

Law  III. — To  every  action  there  is  always  an 
equal  and  contrary  reaction;  or,  the  mutual  ac- 
tions of  any  two  bodies  are  always  equal  and  oppo- 
sitely directed. 

167.  Remarks   on  Law  I.— Law  I  supplies  us  with  a 

definition  of  force.  It  indicates  that  force  is  that  which  tends  to 
change  a  body's  state  of  rest  or  of  uniform  motion  in  a  straight  line  ; 
for  if  a  body  does  not  continue  in  its  state  of  rest  or  of  uniform  mo- 
tion in  a  straight  line  it  must  be  under  the  action  of  force. 

A  body  has  no  power  to  change  its  own  state  as  to  rest  or  motion  ; 
when  it  is  at  rest,  it  has  no  power  of  putting  itself  in  motion  ;  when 
in  motion  it  has  no  power  of  increasing  or  diminishing  its  velocity. 
Matter  is  inert  (Art.  3).  If  it  is  at  rest,  it  will  remain  at  rest ;  if  it  is 
moving  with  a  given  velocity  along  a  rectilinear  path,  it  will  continue 
to  move  with  that  velocity  along  that  path.  It  is  alike  natural  to 
matter  to  be  at  rest  or  in  motion.  Whenever,  therefore,  a  body's 
state  is  changed  either  from  rest  to  motion,  or  from  motion  to  rest, 
or  when  its  velocity  is  increased  or  diminished,  that  change  is  due  t<( 
some  external  cause.  This  cause  is  called  force  (Art.  14) ;  and  the 
word  farce  is  used  in  Kinetics  in  this  meaning  only. 


*  Thomso     lad  Tail's  Nat.  Phil.,  p.  Ml. 


REMARKS    ON  LAW  II.  291 

168.  Remarks  on  Law  II. — Law  II  asserts  that  if  any 
/orce  generates  motion,  a  double  force  will  generate  double  motion 
and  so  on,  whether  applied  simultaneously  or  successively,  instan- 
taneously or  gradually.  And  this  motion,  if  the  body  was  moving 
beforehand,  is  either  added  to  the  previous  motion  if  directly  conspir- 
ing with  it,  or  is  subtracted  if  directly  opposed  ;  or  is  geometrically 
compounded  with  it  according  to  the  principles  already  explained 
(Art,  29),  if  the  line  of  previous  motion  and  the  direction  of  the  force' 
are  inclined  to  each  other  at  an  angle.  The  term  motion,  here  means 
quantity  of  motion,  and  the  phrase  change  of  motion  here  means  rate 
of  change  of  quantity  of  motion  (Art.  13).  If  the  force  be  finite  it  will 
require  a  finite  time  to  produce  a  sensible  change  of  motion,  and  the 
change  of  momentum  produced  by  it  will  depend  upon  the  time. dur- 
ing which  it  acts.  The  change  of  motion  must  then  be  understood  ta 
be  the  change  of  momentum  produced  per  unit  of  time,  or  the  rati 
of  change  of  momentum,  or  acceleration  of  momentum,  which  agrees 
with  the  principles  already  explained  (Arts.  13  and  20).  In  this  law 
nothing  is  said  about  the  actual  motion  of  the  body  before  it  was 
acted  on  by  the  force;  it  is  only  the  change  of  motion  that  concerns 
us.  The  same  force  will  produce  precisely  the  same  change  of  mo- 
tion in  a  body  ;  whether  the  body  be  at  rest,  or  in  motion  with  any 
velocity  whatever. 

Since,  when  several  forces  act  at  once  on  a  particle  either 
at  rest  or  in  motion,  the  second  law  of  motion  is  true  for 
every  one  of  these  forces,  it  follows  that  each  must  have  the 
same  effect,  in  so  far  as  the  change  of  motion  produced  by 
it  is  concerned,  as  if  it  were  the  only  force  in  action. 
Hence  the  assertion  of  the  second  law  may  be  put  in  the 
following  form : 

When  any  number  of  forces  act  simultaneously  on  a  body, 
whether  at  rest  or  in  motion  in  any  direction,  each  force  pro- 
duces  in  the  body  the  same  change  of  motion  as  if  it  alone 
had  acted  on  the  body  at  rest. 

It  follows  from  this  view  of  the  law  that  all  problems 
which  involve  forces  acting  simultaneously  may  be  treated 
as  if  the  forces  acted  successively. 

The  operations  of  this  law  have  p1  ready  been  considered  in  Kine 


292  REMARKS   ON  LAW  IT. 

matics  (Art.  149) ;  but  motion  there  was  understood  to  mean  velocity 
ouly,  since  the  mass  of  the  body  was  not  considered.  This  law  in- 
cludes, therefore,  the  law  of  the  composition  of  velocities  already 
referred  to  (Art.  29).  Another  consequence  of  the  law  is  the  follow- 
ing :  Since  forces  are  measured  by  the  changes  of  motion  they  pro- 
duce, and  their  directions  assigned  by  the  directions  in  which  these 
changes  are  produced,  and  since  the  changes  of  motion  of  one  and  the 
same  body  are  in  the  directions  of,  and  proportional  to,  the  changes 
of  velocity,  therefore  a  single  force,  measured  by  the  resultant  change 
of  velocity,  and  in  its  direction,  will  be  the  equivalent  of  any  number 
of  simultaneously  acting  forces. 

Hence, 

The  resultant  of  any  number  of  concurring  forces  is  to  be 
found  by  the  same  geometric  process  as  the  resultant  of  any 
number  of  simultaneous  velocities,  and  conversely. 

From  this  follows  at  once  the  Polygon  of  Velocities  and 
the  Parallelojnjjed  of  Velocities  from  the  Polygon  and 
Parallelopiped  of  Forces,  as  was  described  in  Art.  142, 

This  law  also  gives  us  the  means  of  measuring  force,  and  also  of 
measuring  the  mass  of  a  body :-  for  the  actions  of  different  forces  upon 
the  same  body  for  equal  times,  evidently  produce  changes  of  velocity 
which  are  proportional  to  the  forces.  Also,  if  equal  forces  act  on  dif- 
ferent bodies  for  equal  times,  the  changes  of  velocity  produced  must 
be  inversely  as  the  masses  of  the  bodies.  Again,  if  different  bodies, 
each  acted  on  by  a  force,  acquire  in  the  same  time  the  same  changes 
of  velocity,  the  forces  must  be  proportional  to  the  masses  of  the 
bodies.  This  means  of  measuring  force  is  practically  the  same  as 
that  already  deduced  by  abstract  reasoning  (Arts.  19  and  20). 

It  appears  from  this  law,  that  every  theorem  of  Kine- 
matics connected  with  acceleration  has  its  counterpart  in 
Kinetics.  Thus,  the  measure  of  acceleration  or  velocity 
increment,  (Art.  9),  which  was  discussed  in  Chap.  I  (Arts. 
8   and   9),   and   in   Kinematics  (Art.  135),   and   which  is 

(Ps 
denoted  by  /  or  its  equal  -j-2,  is  also  the  effect  and  the 

measure  of  force ;  therefore  all  the  results  of  the  equation 


REMARKS   ON  LAW  II.  293 

/-£•  « 

its  vanous  forms,  and  the  remarks  which  have  been  made 
on  it,  are  applicable  to  it  when  /  is  the  accelerating  force. 
Thus,  (Art.  1G2),  we  see  that  the  force,  under  which  a 
particle  describes  any  curve,  may  be  resolved  into  two 
components,  one  in  the  tangent  to  the  curve,  the  other 
towards  the  centre  of  curvature;  their  magnitudes  being 
the  acceleration  of  momentum,  and  the  product  of  the 
momentum  into  the  angular  velocity  about  the  centre  of 
curvature,  respectively.  In  the  case  of  uniform  motion, 
the  first  of  these  vanishes,  or  the  whole  force  is  perpen- 
dicular to  the  direction  of  motion.  When  there  is  no  force 
perpendicular  to  the  direction  of  motion,  there  is  no  curva- 
ture, or  the  path  is  a  straight  line. 

Hence  if  we  suppose  the  particle  of  mass  m  to  be  at  the 
point  (x,  y,  z),  and  resolve  the  forces  acting  on  it  into  the 
three  rectangular  components,  X,  Y,  Z,  we  have 

d?x       v         d2y       v        d2z       „  ,_. 

In  several  of  the  chapters  these  equations  will  be  sim- 
plified by  assuming  unity  as  the  mass  of  the  moving 
particle.  When  this  cannot  be  done,  it  is  sometimes  con- 
venient to  assume  X,  Y,  Z,  as  the  component  forces  on  the 
unit  mass,  and  (2)  becomes 

m-j-r  =  mX,  etc. 
at4 

from  which  m  may  of  course  be  omitted.  It  will  be  ob- 
served that  an  equation  such  as 

d?x       v 

w=x 

may  be  interpreted  either  as  Kinetical  or  Kinematical ;  if 


294  REMARKS   ON  LAW  III. 

the  former,  the  unit  of  mass  must  be  understood  as  a  fac- 
tor on  the  left- hand  side,  in  which  case  X  is  the  ^-com- 
ponent, for  the  unit  of  mass,  of  the  whole  force  exerted  on 
the  moving  body. 

The  first  two  laws,  have,  therefore,  furnished  us  with  a  definition 
and  a  measure  of  force;  and  they  also  show  how  to  compound,  and 
therefore  how  to  resolve,  forces ;  and  also  how  to  investigate  the 
conditions  of  equilibrium  or  motion  of  a  single  particle  subjected  to 
given  forces. 

169.    Remarks  on  Law  III. — According  to  Law  III,  if  one 

body  presses  or  draws  another,  it  is  pressed  or  drawn  by  this  other 
with  an  equal  force  in  the  opposite  direction  (Art.  10).  A  horse 
towing  a  boat  on  a  canal,  is  pulled  backwards  by  a  force  equal' to  that 
which  he  impresses  on  the  towing-rope  forwards.  If  one  body  strikes 
another  body  and  changes  the  motion  of  the  other  body,  its  own 
motion  will  be  changed  in  an  equal  quantity  and  in  the  opposite 
direction ;  for  at  each  instant  during  the  impact  the  bodies  exert  on 
each  other  equal  and  opposite  pressures,  and  the  momentum  that  one 
body  loses  is  equal  to  that  which  the  other  gains. 

The  earth  attracts  a  falling  pebble  with  a  certain  force,  while  the 
pebble  attracts  the  earth  with  an  equal  force.  The  result  is  that 
while  the  pebble  moves  towards  the  earth  on  account  of  its  attrac- 
tion, the  earth  also  moves  towards  the  pebble  under  the  influence  of 
the  attraction  of  the  latter  ;  but  the  mass  of  the  earth  being  enor- 
mously greater  than  that  of  the  pebble  while  the  forces  on  the  two 
arising  from  their  mutual  attractions  are  equal,  the  motion  produced 
thereby  in  the  earth  is  almost  incomparably  less  than  that  produced 
in  the  pebble,  and  is  consequently  insensible. 

It  follows  that  the  sum  of  the  quantities  of  motion  parallel  to  any 
fixed  direction  of  the  particles  of  any  system  influencing  one  another 
in  any  possible  way,  remains  unchanged  by  their  mutual  action. 
Therefore  if  the  centre  of  gravity  of  any  system  of  mutually 
influencing  particles  is  in  motion,  it  continues  moving  uniformly  in  a 
straight  line,  unless  in  so  far  as  the  direction  or  velocity  of  its  motion 
is  changed  by  forces  between  the  particles  and  some  other  matter  not 
belonging  to  the  system  ;  also  the  centre  of  gravity  of  any  system  of 
particles  moves  just  as  all  the  matter  of  the  system,  if  concentrated  in 
a  point,  would  move  under  the  influence  of  forces  equal  and  parallel 
to   the   forces   really  acting  on    its    different    parts.      (For    further 


TWO   LA  WS    OF  MOTION.  295 

remarks  on  these  laws  see  Tait  and  Steele's  Dynamics  of  a  Particle, 
Thomson  and  Tait's  Nat.  Phil.,  Pratt's  Mechanics,  etc.) 

170.  Two  Laws  of  Motion  in  the  French  Trea- 
tises.— Newton's  Laws  of  motion  are  not  adopted  in  the 
principal  French  treatises  ;  but  we  find  in  them  tivo  prin- 
ciples only  as  borrowed  from  experience,  viz.: 

First. — The  Law  of  Inertia,  that  a  body,  not  acted 
upon  by  any  force,  would  go  on  for  ever  with  a  uniform 
velocity.     This  coincides  with  Newton's  First  Law. 

Second. — That  the  velocity  communicated  is  proportional 
to  the  force.  The  second  and  third  Laws  of  Motion  are 
thus  reduced  to  this  second  principle  by  the  French  writers, 
especially  Poisson  and  Laplace.* 

171.  Motion  of  a  Particle  under  the  Action  of  an 
Attractive  Force. — A  particle  moves  under  a  force  of 
attraction  which  is  in  its  line  of  motion,  and  varies  directly 
as  the  distance  of  the  particle  from  the  centre  of  force;  it  is 
required,  to  determine  the  motion. 

The  point  whence  the  influence  of  a  force  emanates  is 
called  the  centre  of  force  ;  and  the  force  is  called  an  attrac- 
tive or  a  repulsive  force  according  as  it  attracts  or  repels. 

Let  0  be  the  centre  of  force,  P  the       A, 
position  of  the  particle  at  any  time,  t,  v  _. '  8Q  £x 

its  velocity  at  that  time,  and  let  OP  =  x, 
and  OA  =  a,  where  A  is  the  position  of  the  particle  when 
t  =  0  ;  let  ^  z=  the  absolute  force ;  that  is,  the  force  of 
attraction  on  a  unit  of  mass  at  a  unit's  distance  from  0, 
which  is  supposed  to  be  known,  and  is  sometimes  called 
the  strength  of  the  attraction.     At  present  we  shall  suppose 

*  Parkinson's  Mechanics,  p.  187.  See  paper  by  Dr.  Whewell  on  the  principles 
of  Dynamics,  particularly  as  stated  by  French  writers,  in  the  Edinburgh  Journal  o/ 
Science,  Vol.  VIII. 


296  A    VARIABLE  ATTRACTIVE  FORCE. 

the  mass  of  the  particle  to  be  unity,  as  it  simplifies  the 
equations.  Then  fix  is  the  magnitude  of  the  force  at  the 
distance  x  on  the  particle  of  unit  mass,  or  it  is  the  accelera- 
tion at  P  ;  and  the  equation  of  motion  is 

df2=~^  (1) 

the  negative  sign  being  taken  because  the  tendency  of  the 
force  is  to  diminish  x ; 

2dx  d2x  n       , 

Integrating,  we  get 

g=P(«2-^),  (2) 

if  the  particle  be  at  rest  when  x  =  a  and  t  =  0, 
—  dx 


Va2  —  x* 


=  \£dt9 


the  negative  sign  being  taken,  because  x  decreases  as  t 
increases.  Integrating  again  between  the  limits  correspond- 
ing to  t  =  t  and  t  =  0, 

cos-1  -  =  u  $L 
a 

.•.    t  =  -rcos-1—  (3) 

fi*  a 

From  (2)  it  appears  that  the  velocity  of  the  particle  is 
zero  when  x  =  a  and  —  a  ;  and  is  a  maximum,  viz.:  afi*; 
when  x  =  0.  Hence  the  particle  moves  from  rest  at  A ;  its 
velocity  increases  until  it  reaches  0  where   it  becomes  a 


A    VARIABLE  ATTRACTIVE  FORCE.  297 

maximum,  and  where  the  force  is  zero ;  the  particle  passes 
through  that  point,  and  its  velocity  decreases,  and  at  A',  at 
a  distance  z=  —  a,  becomes  zero.  From  this  point  it  will 
return,  under  the  action  of  the  force,  to  its  original  posi- 
tion, and  continually  oscillate  over  the  space  2a,  of  which 
0  is  the  middle  point. 

From  (3)  we  find  when  x  =  a,  t  =  0  and  when  x  =  0, 

77"  77 

t  =  — r ;  so  that  the  time  of  passing  from  A  to  0  =  — T , 

and  the  time  from  0  to  A'  is  the  same,  so  that  the  time  of 

oscillation  from  A  to  A'  is  -y-     This  result  is  remarkable, 

as  it  shows  that  the  time  of  oscillation  is  independent  of 
the  velocity  and  distance  of  projection,  and  depends  solely 
on  the  strength  of  the  attraction,  and  is  greater  as  that  is 
less. 

This  problem  includes  the  motion  of  a  particle  within  a 
homogeneous  sphere  of  ordinary  matter  in  a  straight  shaft 
through  the  centre.  For  the  attraction  of  such  a  sphere  on 
a  particle  within  its  bounding  surface  varies  directly  as  the 
distance  from  the  centre  of  the  sphere  (Art.  133a).  If  the 
earth  were  such  a  homogeneous  sphere,  and  if  AOA'  (Fig. 
80)  represented  a  shaft  running  straight  through  its  centre 
from  surface  to  surface,  then,  if  a  particle  were  free  at  one 
3iid,  A,  it  would  move  to  the  centre  of  the  earth,  0,  where 
its  velocity  would  be  a  maximum,  and  thence  on  to  the 
opposite  side  of  the  earth,  A',  where  it  would  come  to  rest ; 
then  it  would  return  through  the  centre,  0,  to  the  side,  A, 
from  where  it  started  ;  and  its  motion  would  continue  to  be 
oscillatory,  and  thus  it  would  move  backwards  and  forwards 
from  one  side  of  the  earth's  surface  to  the  other,  and  the 
time  of  the  oscillation  would  be  independent  of  the  earth's 
radius;  that  is,  at  whatever  point  within  the  earth's  surface 
the  particle  be  placed  it  would  reach  the  centre  in  the 
same  time. 


298  A    VARIABLE  REPULSIVE  FORCE. 

Cou. —  To  find  this  time.  Since  \i  is  the  attraction  at  a 
unit  of  distance  and  g  the  attraction  at  the  distance  R}  we 

have  11  =  -^,  which  in  t  =  — T  gives 

for  the  time  it  would  take  a  body  to  move  from  any  point 
within  the  earth's  surface  to  the  centre. 
If  we  put  g  =  32|  feet  and  R  =  3963  miles  we  get 

t  =  21  m.  6  s.  about, 

which  would  be  the  time  occupied  in  passing  to  the  earth's 
centre,  however  near  to  it  the  body  might  be  placed,  or 
however  far,  so  long  as  it  is  within  the  surface. 

172.  Motion  of  a  Particle  under  the  Action  of  a 
Variable  Repulsive  Force. — Let  the  force,  be  one  of 
repulsion  and  vary  as  the  distance,  then  the  equation  of 
motion  is 

dl>=^ 

Let  us  suppose  the  particle  to  be  projected  from  the  cen- 
tre of  force  with  the  velocity  v0  ;  then  we  have 

^2  =  ^2  +  V;  (i) 

2p$ 

As  t  increases  x  also  increases,  and  the  particle  recedes 
further  and  further  from  the  centre  of  force;  and  the 
velocity  also  increases,  and  ultimately  equals  go  when  x  = 
t  =  co .     Thus  in  this  case  the  motion  is  not  oscillatory. 


A    VARIABLE  ATTRACTIVE   FORCE.  299 

173.  Motion  of  a  Particle  under  the  Action  of  an 
Attractive  Force  -which  is  in  the  line  of  motion,  and 
which  varies  Inversely  as  the  Square  of  the  Distance 
from  the  Centre  of  Force. 

Let  0  (Fig.  80)  be  the  centre  of  force,  P  the  position  of 
the  particle  at  the  time  t\  and  A  the  position  at  rest  when 
/  =  0,  so  that  the  particle  starts  from  A  and  moves  to- 
wards 0.  Let  OP  =  x,  OA  =  a,  and  \i  =  the  absolute 
force  as  before  or  the  acceleration  at  unit  distance  from  0. 
Then  the  equation  of  motion  is 

d?x  _        (i 
di?  ~~  ~x* 

Multiplying  by  2dx  and  integrating,  we  get 


dx2 
dP 


whicn  gives  the  velocity  of  the  particle  at  any  distance,  x, 
from  the  origin. 
From  (1)  we  have 

dx  l%i  Vax  —  a? 


dt 


v/f 


the  negative   sign  being  taken  because  in  the  motion  to- 
wards 0,  x  diminishes  as  t  increases.     This  gives 


%\i  —  xdx 

dt 


a  Vax  —  x2 

r.     a  —  2x         a         1  1   , 

=     £      =  —  5  ——  dx. 

L    V  ax  —  x2       *  v  ax  —  x2j 


300  VELOCITY  IN  FALLING* 

Integrating  and  taking  the  limits  corresponding  to  t  =  i 
and  t  =  0,  we  have 


(2) 


which  gives  the  value  of  £. 

When  the  particle  arrives  at  0,  x  =  0,  therefore  the 
time  of  falling  to  the  centre  0  from  A  is 

From  (1)  we  see  that  the  velocity  =  0  when  x  =  a;  and 
=  cc  when  x  =  0 ;  hence  the  velocity  increases  as  the 
particle  approaches  the  centre  of  force,  and  ultimately, 
when  it  arrives  at  the  centre,  becomes  infinite.  And 
although  at  any  point  very  near  to  0  there  is  a  very  great 
attraction  tending  towards  0,  at  the  point  0  itself  there  is 
no  attraction  at  all;  therefore  the  particle,  approaching 
the  centre  with  an  indefinitely  great  velocity,  must  pass 
through  it.  Also,  everything  being  the  same  at  equal 
distances  on  either  side  of  the  centre,  we  see  that  the 
motion  must  be  retarded  as  rapidly  as  it  was  accelerated, 
and  therefore  the  particle  will  proceed  to  a  point  A'  at  a 
distance  on  the  other  side  of  0  equal  to  that  from  which  it 
started  ;  and  the  motion  will  continue  oscillatory. 

174.  Velocity  acquired  in  Falling  through  a  Great 
Height  above  the  Earth. — The  preceding  case  of  motion 
includes  that  of  a  body  falling  from  a  great  height  above 
the  earth's  surface  towards  its  centre,  the  distance  through 
which  it  falls  being  so  great  that  the  variations  of  the  earth's 
attraction  due  to  the  distance  must  be  taken  into  account. 
For  a  sphere  attracts  an  external  particle  with  a  force  which 
varies  inversely  as  the  square  of  the  distance  of  the  particle 


VELOCITY  IN  FALLING.  301 

from  the  centre  of  the  sphere  (Art.  133a);  therefore  if  R  is 
the  earth's  radius,  g  the  kinetic  measure  of  gravity  on  a 
unit  of  mass  at  the  earth's  surface  (Arts.  20,  23),  and  x  the 
distance  of  a  body  from  the  centre  of  the  earth  at  the  time 
t,  then  the  equation  of  motion  is 

®x  _      _     R? 
dP  ~       9  x2 ' 

which  is  the  same  as  the  equation  in  Art.  173  by  writing  ix 
for  gR2;  therefore  the  results  of  the  last  Art.  will  apply  to 
this  case.  Substituting  gR2  for  fi  in  (1)  of  Art.  173  we 
have 

When  the  body  reaches  the  earth's  surface,  x  ==  R  and 
(1)  becomes 

If  a  is  infinite  (2)  becomes 

v  =  VfyR> 

so  that  the  velocity  can  never  be  so  great  as  this,  however 
far  the  body  may  fall;  and  hence  if  it  wflre  possible  to 
project  a  body  vertically  upwards  with  this  velocity  it  would 
go  on  to  infinity  and  never  stop,  supposing,  of  course,  that 
there  is  no  resisting  medium  nor  other  disturbing  force. 
If  in  (2)  we  put  g  =  32£  feet  and  R  =  3963  miles  we 

get 

v  =  [2. 32|-3963. 5280]*  feet  =  6-95  miles; 

bo  that  the  greatest  possible  velocity  which  a  body  can 
acquire  in  falling  to  the  earth  is  less  than  7  miles  per 
second,   and  if  a  body  were  projected  upwards  with  that 


302  MOTION  IN  A    RESISTING   MEDIUM. 

velocity,    and    were    to   meet   with    no   resistance   except 
gravity,  it  would  never  return  to  the  earth. 

Cok. — To  find  the  velocity  which  a  body  would  acquire 
in  falling  to  the  earth's  surface  from  a  height  h  above  the 
surface,  we  have  from  (1)  by  putting  x  =  R  and  a  =  h-\  R, 


"-^'S-itJ 


2(/Rh 


R  +  h 

If  h  be  small  compared  with  R,  this  may  be  written 

v2  =  2gh, 

which  agrees  with  (6)  of  Art,  140. 

The  laws  of  force,  enumerated  in  Arts.  171,  173,  are  the 
only  laws  that  are  known  to  exist  in  the  universe  (Pratt's 
Mechs.,  p.  212). 

175.  Motion  in  a  Resisting  Medium. — In  the  pre- 
ceding discussion  no  account  is  taken  of  the  atmospheric 
resistance.  We  shall  now  consider  the  motion  of  a  body 
near  the  surface  of  the  earth,  taking  into  account  the 
resistance  of  the  air,  which  we  may  assume  varies  as  the 
square  of  the  velocity. 

A  particle  under  the  action  of  gravity,  as  a  constant  force, 
moves  in  the  air  supposed  to  be  a  resisting  medium  of 
uniform  density,  of  which  the  resistance  varies  as  the  square 
of  the  velocity  required  to  determine  the  motion. 

Suppose  the  particle  to  descend  towards  the  earth  from 
rest.  Take  the  origin  at  the  starting  point,  let  the  line  of 
its  motion  be  the  axis  of  x  ;  and  let  x  be  the  distance  of 
the  particle  from  the  origin  at  the  time  t,  and  for  con- 
venience let  gk2  be  the  resistance  of  the  air  on  the  particle 
for  a  unit  of  velocity;  gk2  ii  cnlled  the  coefficient  of  resist- 
ance.    Then  the  resistance  of  the  air  at  the  distance  x  from 


MOTION  IN  A    RESISTING   MEDIUM.  303 

j- J ,  which  acts  upwards,  and  the  force  of 

gravity   is  g   acting   downwards,  the   mass   being  a  unit. 
Kence  the  equation  of  motion  is 

d?x  72  /dx\* 

7J  dt 

.«.    gdt  =  — 


Integrating,  remembering  that  when  t  =  0,  v  =  0,  we 
get 

-    ,   ,  dx 
1         1  +  kdt 
gt  =  U  l0g ^'  (Calculus,  p.  259,  Ex.  5). 

Passing  to  exponentials  we  have 

dx       1  eW  —  e-*0* 


dt  ~    h  eW  +  e-W ' 


(2) 


which  gives  the  velocity  in  terms  of  the  time.     To  find  it  in 
terms  of  the  space,  we  have  from  (1) 

=  2gk2dx; 


'-"§) 


.-.    log[l-Ff|)2J--2y^,  (3) 

observing  the  proper  limits ; 


304  MOTION   OF  ASCENT  IN  THE  AIR. 

...  £-£a-^"*  w 

which  gives  the  velocity  in  terms  of  the  distance. 

Also,  integrating  (2)  taking  the  same  limits  as  before, 
we  get 

gk*x  =  log  (eW  +  e-W)  —  log  2  ; 

.*.     2effk*x  =  eW  +  e-W,  (5) 

which  gives  the  relation  between  the  distance  and  the  time 
of  falling  through  it. 

As  the  time  increases  the  term  e-W  diminishes  and  from 
(5)  the  space  increases,  becoming  infinite  when  the  time  is 
infinite;  but  from  (2),  as  the  time  increases  the  velocity 
becomes   more    nearly   uniform,    and   when    t  =  go,    the 

velocity  =  j-;  and  although  this  state  is  never  reached,  yet 

it  is  that  to  which  the  motion  approaches. 

176.  Motion  of  a  Particle  Ascending  in  the  Air 
against  the  Action  of  Gravity. — Let  us  suppose  the 
particle  to  be  projected  upwards,  that  is,  in  a  direction 
contrary  to  that  of  the  action  of  gravity,  with  a  given 
velocity,  v,  it  is  required  to  determine  the  motion. 

Let  us  suppose  the  particle  to  be  of  the  same  form  and 
size  as  before,  and  the  same  coefficient  of  resistance. 
Then,  taking  x  positive  upwards,  both  gravity  and  the 
resistance  of  the  air  tend  to  diminish  the  velocity  as  t 
increases ;  so  that  the  equation  of  motion  is 

cPx  7JdA2  „v 

&  =  -*-&\3ih  (1) 


MOTION   CF  ASCENT  IN   THE  AIR.  305 


•Jr 

=  —kg 

tft; 

•   • 

•'♦•ST 

.*.    tan 

1  A;  -TT  =  tan 
at 

*(*»)- 

-#*; 

(Calculus,  p.  244,  Ex.  3),  since  the  initial  velocity  is  v. 

Taking  the  tangent  of  both  members  and  solving  for 

dx 

m,  we  get 

dx  _  1      v&  —  tan  hgt  #  ,  . 

5?  ="  k'l  +  vhtmhgi'  (' 

which  gives  the  velocity  in  terms  of  the  time.     To  find  it 
in  terms  of  the  distance,  we  have  from  (1) 


-er 


*+*© 


SB?  =-*•'** 


<fe\» 


•••    bg      1  +  5.     =      -****  (3) 

.• .     (^)2  =  &e-*g»j*  -  1  (1  -  fr-^*,»),  (4) 

which  gives  the  velocity  in  terms  of  the  distance. 

Also,  integrating  (2)  after  substituting  sine  and  cosine 
for  tangent,  and  taking  the  same  limits  as  before,  we  get 

gj£x  =  log  (vh  sin  hgt  +  cos  hgt) ;  (5) 

which  gives  the  space  described  by  the  particle  in  terms  of 
the  time. 


306  MOTION  OF  ASCENT  IN  THE  AIR. 

Cor.   1. — To  find  the  greatest  height  to  which  the  par- 

clx 
tide  will  ascend  put  the  velocity,  -y  =  0,  in  (3)  and  get 

x  =  p* log  (1  +  W)>  <6> 

which  is  the  distance  of  the  highest  point. 

floe 
Putting  -jT  =  0  in  (2)  we  get 

t  =  j-  tan-1  vh,  (7) 

which  is  the  time  required  for  the  particle  to  reach  the 
highest  point.  Having  reached  the  greatest  height,  the 
particle  will  begin  to  fall,  and  the  circumstances  of  the 
fall  will  be  given  by  the  equations  of  Art.  175. 

Cor.  2. — Since  h  is  the  same  in  this  and  Art.  175,  we 
may  compare  the  velocity  of  projection,  v,  with  that  which 
the  particle  would  acquire  in  descending  to  the  point 
whence  it  was  projected.  Denote  by  v0  the  velocity  of 
the  particle  when  it  reaches  the  point  of  starting.  From 
(3)  of  Art.  175  we  have 

X  =  2jj&  l0g  1  -  l*v0*> 

and  placing  this  value  of  x  equal  to  that  given  in  (6), 
we  get, 


vn  = 


(1  +  &v*)*' 
which  is  less  than  v ;    hence   the  velocity  acquired  in  the 


MOTION   OF  A    PROJECTILE.  307 

descent  is  less  than  that  lost  in  the  ascent,  as  might  have 
heen  inferred. 

Cor.  3. — Substituting  (G)  in  (5)  of  Art.  175,  we  get  for 
the  time  of  the  descent, 


t  =  j~  log  ( VI  +  PP  +  lev), 

which  is  different  from  the  time  of  the  ascent  as  given  in 
(7).  (See  Price's  Anal.  Mech's,  Vol.  I,  p.  40G ;  Venturoli's 
Mech's,  p.  82  ;  Tait  and  Steele's  Dynamics  of  a  Particle, 
p.  237.) 

177.  Motion  of  a  Projectile  in  a  Resisting  Me- 
dium.— The  theory  of  the  motion  of  projectiles  in  vacuo, 
which  was  examined  under  the  head  of  Kinematics,  affords 
results  which  differ  greatly  from  those  obtained  by  direct 
experiment  in  the  atmosphere.  When  projectiles  move 
with  but  small  velocity,  the  discrepancy  between  the  para- 
bolic theory,  and  what  is  found  to  occur  in  practice,  is 
small  ;  but  with  increasing  velocities,  as  those  with  which 
balls  and  shells  traverse  their  paths,  the  air's  resistance 
increases  in  a  higher  ratio  than  the  velocity,  so  that  the 
discrepancy  becomes  very  great. 

The  most  important  application  of  the  theory  of  projec- 
tiles, is  that  of  Gunnery,  in  which  the  motion  takes  place 
in  the  air.  If  it  were  allowable  to  neglect  the  resistance  of 
the  air  the  investigations  in  Part  II  would  explain  the 
theory  of  gunnery  ;  but  when  the  velocity  is  considerable, 
the  atmospheric  resistance  changes  the  nature  of  the  tra- 
jectory so  much  as  to  render  the  conclusions  drawn  from 
the  theory  of  projectiles  in  vacuo  almost  entirely  inap- 
plicable in  practice. 

The  problem  of  gunnery  may  be  stated  as  follows: 
Given  a  projectile  of  known  wreight  and  dimensions, 
starting  with  a  known  velocity  at  a  known  angle  of  eleva- 


308  MOTIOX  OF  A    PROJECTILE. 

tion  in  a  calm  atmosphere  of  approximately  known  density; 
to  find  its  range,  time  of  flight,  velocity,  direction,  and 
position,  at  any  moment ;  or,  in  other  words,  to  construct 
its  trajectory.  This  problem  is  not  yet,  however,  suscepti- 
ble of  rigorous  treatment ;  mathematics  has  hitherto  proved 
unable  to  furnish  complete  formulae  satisfying  the  condi- 
tions. The  resistance  of  the  air  to  slow  movements,  say  of 
10  feet  per  second,  seems  to  vary  with  the  first  power  of 
the  velocity.  Above  this  the  ratio  increases,  and  as  in  the 
case  of  the  wind,  is  usually  reckoned  to  vary  as  the  square 
of  the  velocity;  beyond  this  it  increases  still  further,  till  at 
1200  feet  per  second  the  resistance  is  found  to  vary  as  the 
cube  of  the  velocity.  The  ratio  of  increase  after  this  point 
is  passed  is  supposed  to  diminish  again ;  but  thoroughly 
satisfactory  data  for  its  determination  do  not  exist. 

From  experiments*  made  to  determine  the  motion  of 
cannon-balls,  it  appears  that  when  the  initial  velocity  is 
considerable,  the  resistance  of  the  air  is  more  than  20  times 
as  great  as  the  weight  of  the  ball,  and  the  horizontal  range 
is  often  a  small  fraction  of  that  which  the  theory  of  pro- 
jectiles in  vacuo  gives,  so  that  the  form  of  the  trajectory  is 
very  different  from  that  of  a  parabolic  path.  Such  experi- 
ments have  been  made  wTith  great  care,  and  show  how  little 
the  parabolic  theory  is  to  be  depended  upon  in  determining 
the  motion  of  military  projectiles. 

178.  Motion  of  a  Projectile  in  the  Atmosphere 
Supposing  its  Resistance  to  vary  as  the  Square  of 
the  Velocity. — A  particle  under  the  action  of  gravity  is 
projected  from  a  given  point  in  a  given  direction  with  a 
given  velocity,  and  moves  in  the  atmosphere  ivhose  resistance 
is  assumed  to  vary  as  the  square  of  the  velocity  ;  to  deter- 
mine the  motion. 


*  See  Encyclopoedia  Britannica,  Art.  Gunnery ;   also  Robin's  Gunnery,  and 
Hutton's  Tracts. 


MOTION  OF  A    PROJECTILE.  309 

Take  the  given  point  as  origin,  the  axis  of  x  horizontal, 
the  axis  of  y  vertical  and  positive  upwards,  so  that  the 
direction  of  projection  may  be  in  the  plane  of  xy.  Let  v 
be  the  velocity  of  projection,  g  the  acceleration  of  gravity, 
«  the  angle  between  the  axis  of  x  and  the  line  of  projection, 
and  let  the  resistance  of  the  air  on  the  particle  be  k  for  a 
unit  of  velocity ;  then  the  resistance,  at  any  time,  t,  in  the 

(ds\2 
-r\  ;  and  the  x-  and  ^-components  of 

this  resistance  are,  respectively, 

,  c Is    dx  1     ,  ds    dy 

^di'di'   ancl  ^di'dt' 

Then  the  equations  of  motion  are,  resolving  horizontally 
and  vertically, 

d2x  ds    dx  .  . 

dP  ~  "     It    di'  W 


(2) 


d?y 
dP  — 

(1)  we  have 

— 7 —  =  —  Ms ; 

dx 

dx 

.          di 
...     log = 

°  v  cos  a 

di 

=  —  ks; 


dx 
since  when  t  =  0,  ^7  =  v  cos  « ; 


dx 

di 


=  v  cos  a  e-*8.  (3) 


Multiplying  (1)  and  (2)  by  dy  and  dx,  respectively,  and 
subtracting  the  former  from  the  latter  we  have 

d2y  dx  —  d*x  dy 

-dfi— *  =  -!**•  W 


310  MOTIOX  OF  A   PROJECTILE. 

Substituting  in  (4)  for  dt2  its  value  from  (3)  we  get 

d2i/dx  —  d?xdy         1dv  q         _.   , 

dxl  dx  v2  cos4 «  v  ' 

Substituting  in  the  second  member  of  (5)  for  dx  its  value 

ds  +  yi  +  Jj£,  we  get 

d4(i  +  jQ*=._JL^^fc  (6) 

t/o;  \         dx2/  v2  cos2 «  v  ' 

dn 
Put  ~  =  p9  and  (6)  becomes 

(1  +  irf  dp  =  —  t-^s-  e3**  <fe. 
1,1  v2  cos2  « 

Integrating,  and  remembering  that  when  5  =  0,  jo  =  tan  «, 
we  get 

P  (1  +  ^2)  *   +  log  |>  +   (1   +  /*)*] 

=  *  -  znr^—  e2fe-  (?) 

KV2  COS2  «  v    ' 

where  c  is  the  constant  of  integration  whose  value 

=  tan  «  sec  «  +  log  (tan  a  +  sec  «)  +  7  0       0    •    (8) 
°  v  '       W  cos2  a     v  ' 

From  (5)  we  have 

z'2  cos2  cc  efa  Way ' 

which  in  (7)  gives 

p(l  +^  +  log  [p  +  (1  +**)*]  -*  =  l% 

=  Mx,      (9) 


'  P  (1  +  ^  +  log  [jp  +  (1  +  ;^2)i]  -  c 


MOTION   OF  A    PROJECTILE.  311 

and    i **£ J- =  kdy.    (10) 

p(l+^*  +  log[i>  +  (l+tf)*]-* 

From  (4)  we  have 

dx-  dp  —  —  gdt2. 

Substituting  this  value  of  dx  in  (9)  and  solving  for  dt  wo 
get 

tlie  negative  sign  of  c//?  being  taken  because  jt?  is  a  decreas- 
-    ing  function  of  /. 

dii 
Replacing  the  value  of  p  =  --,  (9),  (10),  and  (11)  become 

dx 

d-y- 

1  dx 

dX  =  l  llv  d  +  dA*  4.  w  I  d*  +  A  +  W     , '   (A) 


*2('+»*+  *ffi+fr+ SI 


(B) 


1  (fo 

* = *»*  h»t8,-»flw«+»r <0) 

from  which  equations,  were  it  possible  to  integrate  them, 

x,  y,  and  t  might  be  found  in  terms  of  -j-  ;  and  if  -=-  were 

eliminated  from  the  two  integrals,  of  (A)  and  (B),  the  re- 
sulting equation  in  terms  of  x  and  y  would  be  that  of  the 


312  MOTION  OF  A    PROJECTILE. 

required  trajectory.  But  these  equations  cannot  be  inte« 
grated  in  finite  terms;  only  approximate  solutions  of  them 
can  be  made  ;  and  by  means  of  these  the  path  of  the  pro- 
jectile may  be  constructed  approximately.  (See  Venturoli's 
Mechs.,  p.  92.) 

Squaring  (A)  and  (B),  and  dividing  their  sum  by  the 
square  of  (C)  we  get 

dtf 


ds*       g  *  +  dx* 


(D) 


<ffl       k        dy  (        dtM  vdy      /        df\ii 


which  gives  the  velocity  in  terms  of  ~« 


179.  Motion  of  a  Projectile  in  the  Atmosphere 
under  a  small  Angle  of  Elevation. — The  case  fre- 
quently occurs  in  practice  where  the  angle  of  projection  is 
very  small,  and  where  the  projectile  rises  but  a  very  little 
above  the  horizontal  line.  In  this  case  the  equation  of  the 
part  of  the  trajectory  that  lies  above  the  horizontal  line 
may  easily  be  found ;   for,  the  angle  of  projection  being 

dv 
very  small,  -^  will  be  very  small,  and  therefore,  throughout 

the  path  on   the  upper  side  of  the   axis  of  x,  powers  of 

dy 

-~  higher  than  the  first  may  be  neglected.     In  this  case 

then 

ds  =  dx\       .♦.    8  =s  x\ 

which  in  (5)  of  Art.  178,  becomes 


dd7r  =  --i-3-!-****; 

ax  v2  cos2  « 


EXAMPLES.  313 

Integrating,  we  get 

since  when  x  =  0,  ~  =  tan  «. 
ax 

Integrating  again  we  get 

y  =  x  tan  «  +  ^-^^ ,ya  /    2     (<?**  -  1).  (1) 

47  2kv2  cos2  a       4&V  cos2  a  x  '    v  ' 

Expanding  e2fcB  in  a  series,  (1)  becomes 

y  =  a;  tan  a  —  ^-/ — = ^-f — 5 ....       (2) 

47  2^  cos**  a       ovi  cos^  «  x  ' 

the  first  two  terms  of  which  represent  the  trajectory  in 
vacuo.     [See  (3)  of  Art.  151.] 
From  (3)  of  Art.  178,  we  have 

dt  = dx  • 

vcos  a 

...    t  =  P=-  (3) 

#z;cos«  *  ' 

which  gives  the  time  of  flight  in  terms  of  the  abscissa. 

The  most  complete  and  valuable  series  of  experiments 
on  the  motion  of  projectiles  in  the  atmosphere  that  has  yet 
been  made,  is  that  of  Prof.  F.  Bashforth  at  Woolwich. 

EXAMPLES. 

1.  Find  how  far  a  force  equal  to  the  weight  of  n  lbs., 
would  move  a  weight  of  m  lbs.  in  t  seconds ;  and  find  the 
velocity  acquired. 


314 


EXAMPLES. 


Here  P  = 
25  we  have 


and   W  =  m ;  therefore  from  (1)  of  Art. 


-™f. 


J  -  m' 


which    in 

not 
v  —  -*-; 
m 


(4)   and   (5) 
and 


respectively  of    (Art.     9),     gives 


.  =  *»* 

2  m 


2.  A  body  weighing  n  lbs.  is  moved  by  a  constant  force 
which  generates  in  the  body  in  one  second  a  velocity  of  a 


feet  per  second ;  find  the  force  in  pounds. 


Ans. 


na 


lbs. 


3.  Find  in  what  time  a  force  of  4  lbs.  would  move  a 
weight  of  9  lbs.  through  49  ft.  along  a  smooth  horizontal # 
plane  ;  and  find  the  velocity  acquired. 

Ans.  t  =  — = ;  v  =  kqt. 
V2g' 

4.  Find  the  number  of  inches  through  which  a  force  of 
one  ounce,  constantly  exerted,  will  move  a  mass  weighing 
one  lb.  in  half  a  second.  Ans.  3g  (£)5. 


X 


5.  Two  weights,  P  and  Q,  are  connected  by  a 
which  passes  over  a  smooth  peg  or  pulley  ;  required 
determine  the  motion. 

Since  the  peg  or  pulley  is  perfectly 
smooth  the  tension  of  the  string  is  the 
same  throughout;  hence  the  force  which 
causes  the  motion  is  the  difference  between 
the  weights,  P  and  Q,  the  weight  of  the 
string  being  neglected.  The  moving  force 
therefore  is  P  —  Q ;  but  the  weight  of  the 
mass  moved  is  P  +  Q.  Hence  substituting 
in  (1)  of  Art.  25,  we  get 

P+  Q 


string 


to 


p 


=n 


Fig.  80a. 


P  - 


/; 


EXAMPLES.  315 

••./-.  5=$*  (i) 

which  is  the  acceleration. 

Substituting  this  in  (4)  and  (5)  of  Art  9,  we  have 

v  =  J^g*  (2) 

which  gives  the  velocity  and  space  at  the  time  t,  the  initial 
velocity  v0  being  0. 

6.  A  body  whose  weight  is  Q,  rests  on  a  smooth  hori- 
zontal table  and  is  drawn  along  by  a  weight  P  attached  to 
it  by  a  string  passing  over  a  pulley  at  the  edge  of  the  table ; 
find  the  motion  of  the  bodies. 

Since  the  weight  Q  is  entirely  supported  by  the  resistance 
of  the  table,  the  moving  force  is  the  weight  P,  hanging 
vertically  downwards,  and  the  weight  of  the  mass  moved  is 
P  +  Ql  therefore  from  (1)  we  have 


f=TTH'9  (1) 

and  this  in  (4)  and  (5)  of  Art.  9   gives   the  velocity  and 
space. 

7.  Eequired  the  tension,  T,  of  the  string  in  the  pre- 
ceding example. 

Here  the  tension  is  evidently  that  force  which,  acting 
along  the  string  on  the  body  whose  weight  is  Q,  produces 

P 

in  it  the  acceleration,  -^ -=  g,  and  therefore  is  measured 

by  the  mass  of  Q  into  its  acceleration.     Hence 


316  EXAMPLES. 

9      P+Qg  ~  P+Q 

8.  Find  the  tension,  T,  of  the  string  in  Ex.  5. 

Here  the  tension  equals  the  weight  Q,  plus  the  force 
which,  acting  aloug  the  string  on  Q,  produces  in  it  the 
acceleration 


P+Q9 

f 

-•+?■ 

P-Q 
P+Qg' 

2PQ 

~  P  +  Q' 

or  it  equals  P  minus  the  accelerating  force  which,  of  course, 
gives  the  same  result. 

9.  Two  weights  of  9  lbs.  and  7  lbs.  hang  over  a  pulley,  as 
in  Ex.  5 ;  motion  continues  for  5  sees.,  when  the  string 
breaks;  find  the  height  to  which  the  lighter  weight  will 
rise  after  the  breakage. 

Substituting  in  (2)  of  Ex.  5  we  have 

v  =  -&32.5  =  20; 

therefore  each  weight  has  a  velocity  of  20  feet,  when  the 
string  breaks.  Hence  from  (6)  of  Art.  9,  we  have  (calling 
g  32  ft.) 

*  =  W  =  6}; 

that  is,  the  lighter  weight  will  rise  6}  feet  before  it  begins 
to  descend. 

10.  A  steam  engine  is  moving  on  a  horizontal  plane  at 
the  rate  of  30  miles  an  hour  when  the  steam  is  turned  off ; 
supposing  the  resistance  of  friction  to  be  T^15-  of  the  weight, 
find  how  long  and  how  far  the  engine  will  run  before  it 
stops. 


EXAMPLES.  31? 

Let  W  6e  the  weight  of  the  engine ;  then  the  resistance 

W 

si  friction  is  jz-r,  and  this  is  directly  opposed  to  motion, 

.    JL-E*.      .    /•__!_. 

"  "    400  ""  gJ  '        '  *    J  ~~  400 

pro.       i    •*         •    o^     -i  i.  30x1760x3        ,, 

The  velocity,  v,  is  30  miles  an  hour  =  — — — — —  =  44 
"  60  x  60 

feet  per  second.     Substituting  these  values  of /and  v  in  the 

equation  v  =  ft,  we  get 

.    t  =  550  sees., 

which  is  the  time  it  will  take  to  bring  the  engine  to  rest  if 
the  velocity  be  retarded  -£fo  feet  per  second. 
Also  v2  =  2fs,  therefore 

s  =  44**4**00  =  12100  feet. 

11.  A  man  whose  weight  is  W,  stands  on  the  platform 
of  an  elevator,  as  it  descends  a  vertical  shaft  with  a  uniform 
acceleration  of  \g ;  find  the  pressure  of  the  man  upon  the 
platform. 

Let  P  be  the  pressure  of  the  man  on  the  platform  when 
it  is  moving  with  an  acceleration  of  \g ;  then  the  moving 
force  is  W  —  P;  and  the  weight  moved  is  W ;  therefore 

W-P  =  ^y;       .-.    P  =  $W- 

12.  A  plane  supporting  a  weight  of  12  ozs.  is  descending 
with  a  uniform  acceleration  of  10  ft.  per  second ;  find  the 
pressure  that  the  weight  exerts  on  the  plane. 

Ans.  8 J  ozs. 


318  EXAMPLES. 

13.  A  weight  of  24  lbs.  hanging  over  the  edge  of  a 
smooth  table  drags  a  weight  of  12  lbs.  along  the  table; 
find  (1)  the  acceleration,  and  (2)  the  tension  of  the  string. 

Ans.   (1)  21 J  ft.  per  sec. ;    (2)  8  lbs. 

14.  A  weight  of  8  lbs.  rests  on  a  platform ;  find 
its  pressure  on  the  platform  (1)  if  the  latter  is  de- 
scending with  an  acceleration  of  \g,  and  (2)  if  it  is 
ascending  with  the  same  acceleration. 

Ans.   (1)  7  lbs.;  (2)  9  lbs. 

15.  Two  weights  of  80  and  70  lbs.  hang  over  a  smooth 
pulley  as  in  Ex.  5  ;  find  the  space  through  which  they  will 
move  from  rest  in  3  sees.  Ans.  9|  ft. 

16.  Two  weights  of  15  and  17  ounces  respectively  hang 
over  a  smooth  pulley  as  in  Ex.  5  ;  find  the  space  de- 
scribed and  the  velocity  acquired  in  five  seconds  from  rest. 

Ans.  s  =  25,  v  =  10. 

17.  Two  weights  of  5  lbs.  and  4  lbs.  together  pull  one 
of  7  lbs.  over  a  smooth  fixed  pulley,  by  means  of  a  con- 
necting string;  and  after  descending  through  a  given 
space  the  4  lbs.  weight  is  detached  and  taken  away  without 
interrupting  the  motion  ;  find  through  what  space  the 
remaining  5  lbs.  weight  will  descend. 

Ans.  Through  f  of  the  given  space. 

18.  Two  weights  are  attached  to  the  extremities  of  a 
string  which  is  hung  over  a  smooth  pulley,  and  the  weights 
are  observed  to  move  through  6.4  feet  in  one  second  ;  the 
motion  is  then  stopped,  and  a  weight  of  5  lbs.  is  added 
to  the  smaller  weight,  which  then  descends  through  the 
same  space  as  it  ascended  before  in  the  same  time  ;  deter- 
mine the  original  weights.  Ans.  £  lbs. ;  -^  lbs. 

19.  Find  what  weight  must  be  added  to  the  smaller 
weight  in  Ex.  5,  so  that  the  acceleration  of  the  system  may 


EXAMPLES.  319 

have  the  same  numerical  value  as  before,  but  may  be  in 

the  opposite  direction.                                .        P2  —  Q2 
11  Ans.  ~ 


20.  A  body  is  projected  up  a  rough  inclined  plane  with 
the  velocity  which  would  be  acquired  in  falling  freely 
through  12  feet,  and  just  reaches  the  top  of  the  plane  ; 
the  inclination  of  the  plane  to  the  horizon  is  00°,  and  the 
coefficient  of  friction  is  equal  to  tan  30°;  find  the  height  of 
the  plane.  Ans.  9  feet. 

21.  A  body  is  projected  up  a  rough  inclined  plane  with 
the  velocity  2g  ;  the  inclination  of  the  plane  to  the  horizon 
is  30°,  and  the  coefficient  of  friction  is  equal  to  tan  15° ; 
find  the  distance  along  the  plane  which  the  body  will 
describe.  Ans.  g  (\/3  +  1). 

22.  A  body  is  projected  up  a  rough  inclined  plane ;  the 
inclination  of  the  plane  to  the  horizon  is  a,  and  the  coef- 
ficient of  friction  is  tan  e ;  if  m  be  the  time  of  ascending, 
and  n  the  time  of  descending,  show  that 

/m\2_  sin  (a  —  e) 
\n  I  ~  sin  (a  +  e) 

23.  A  weight  P  is  drawn  up  a  smooth  plane  inclined  at 
an  angle  of  30°  to  the  horizon,  by  means  of  a  weight  Q 
which  descends  vertically,  the  weights  being  connected  by 
a  string  passing  over  a  small  pulley  at  the  top  of  the  plane ; 
if  the  acceleration  be  one-fourth  of  that  of  a  body  falling 
freely,  find  the  ratio  of  Q  to  P.  Ans.   Q  =  P. 

24.  Two  weights  P  and  Q  are  connected  by  a  string, 
and  Q  hanging  over  the  top  of  a  smooth  plane  inclined  at 
30°  to  the  horizon,  can  draw  P  up  the  length  of  the  plane 
in  just  half  the  time  that  P  would  take  to  draw  up  Q\ 
show  that  Q  is  half  as  heavy  again  as  P. 


820  EXAMPLES. 

25.  A  particle  moves  in  a  straight  line  under  the  action 
of  an  attraction  varying  inversely  as  the  (|)th  power  of 
the  distance;  show  that  the  velocity  acquired  by  falling 
from  an  infinite  distance  to  a  distance  a  from  the  centre  is 
equal  to  the  velocity  which  would  be  acquired  in  moving 

from  mst  at  a  distance  a  to  a  distance  -• 

4 


CHAPTER    II. 

CENTRAL    FORCES.* 

180.  Definitions. — A  central  force  is  one  which  acts 
directly  towards  or  from  a  fixed  point,  and  is  called  an 
attractive  or  a  repulsive  force  according  as  its  action  on 
any  particle  is  attraction  or  repulsion.  The  fixed  point  is 
called  the  Centre.  The  intensity  of  the  force  on  any  par- 
ticle is  some  function  of  its  distance  from  the  centre. 
Since  the  case  of  attraction  is  the  most  important  applica- 
tion of  the  subject,  we  shall  take  that  as  our  standard  case ; 
but  it  will  be  seen  that  a  simple  change  of  sign  will  adapt 
our  general  formulas  to  repulsion.  If  the  centre  be  itself 
in  motion,  we  may  treat  it  as  fixed,  in  which  case  the  term 
"actual  motion  "of  any  particle  means  its  motion  "rela- 
tive "  to  the  centre,  taken  as  fixed. 

The  line  from  the  centre  to  the  particle,  is  called  a 
Radius  Vector.  The  path  of  the  particle  under  the  action 
of  an  attraction  or  repulsion  directed  to  the  centre  is 
called  its  Orbit.\  All  the  forces  of  nature  with  which  we 
are  acquainted,  are  central  forces;  for  this  reason,  and  be- 
cause the  motion  of  bodies  under  the  action  of  central 
forces  is  a  branch  of  the  general  theory  of  Astronomy,  we 
shall  devote  this  chapter  to  the  consideration  of  their 
action. 

181.  A  Particle  under  the  Action  of  a  Central 
Attraction;  Required  the  Polar  Equation  of  the 
Path. — The  motion  will  clearly  take  place  in  the  plane 
passing  through  the  centre,  and  the  line  along  which  the 

*  This  chapter  contains  the  first  principles  of  Mathematical  Astronomy.     It 
may,  however,  be  omitted  by  the  student  of  Engineering, 
t  Called  Central  Orbits. 


322 


CEXTR AL   A TTRA  CT10N. 


particle  is  initially  projected,  as  there  is  nothing  to  with- 
draw  the  particle  from  it.  Let  the  centre  of  attraction,  03 
6e  the  origin,  and  OX,  OY,  any 
two  lines  through  0  at  right  angles 
to  each  other,  he  the  axes  of  co- 
ordinates. Let  (x,  y)  be  the 
position  of  the  particle  M  at  the 
time  t,  and  (r,  6)  its  position 
referred  to  polar  co-ordinates, 
OX  being  the  initial  line.  Then, 
calling  P  the  central  attractive 
force,  we  have  for  the  components  parallel  to  the  axes  of  x 

x  1/ 

and  y,  respectively,  —  P-,  —  P-,  the  forces  being  nega- 
tive, since  they  tend  to  diminish  the  co-ordinates.  There- 
fore the  equations  of  motion  are 


x     d2y  _ 


dt* 


w 


Multiplying  the  former  by  y,  and  the  latter  by  x,  and 
subtracting,  we  have 


d*y  _      cPx 
Xdt*       y  dP 

Integrating  we  have 


=  0. 


dy         dx       , 

xl-yTt=h> 

where  h  is  an  undetermined  constant. 
Since  x  =  r  cos  0,  and  y  =  r  sin  6,  we  have 
dx  =  cos  6  dr  —  r  sin  0  dd, 
dy  =  sin  0  dr  +  r  cos  6  dd, 

which  in  (3)  gives 


(2) 
(3) 


(4) 


CENTRAL  ATTRACTION,  323 

Again,  multiplying  the  first  and  second  of  (1)  by  2dx 
and  2dy  respectively,  and  adding,  we  get 

2dx  d2x  +  2dy  d2y  _     _  2P  (x  dx  +  y  dy)  m 
~~dP  ~~  "  r  ' 

-v '*«  +  *)—«*  <6> 

Substituting  in  (6)  the  values  of  tfo2  and  dy2  from  (4),  we 
have 

4(S +")£]=-»"*•= 

,  /l  A-2       1  \  2P  ,    ,     ...  ,„. 

Put  r  =  -  ;  and  .  •.  dr  == 5 :  and  (7)  becomes 

w'  u2'         v  ' 

'^2         A        2P 


,  /rfw>    ,      A         2P 


performing  the  differentiation  of  the  first  member,  and 
dividing  by  2du>  and  transposing,  we  get 

cPu    t  P         n  /Qv 

5P  +  «  -  59  =  °-  (8> 

which  is  the  differential  equation  of  the  orbit  described; 
and  as,  in  any  particular  instance,  the  force  P  will  be  given 
in  terms  of  r,  and  therefore  in  terms  of  u,  the  integral  of 
this  equation  will  be  the  polar  equation  of  the  required 
path. 

Solving  (8)  for  P  we  have 


324  CENTRAL   ATTRACTION. 

J»=*JW  (£  +  .);  (9) 

which  is  the  same  result  that  was  found  by  a  different  pro- 
cess in  Art.  163  for  the  acceleration  along  the  radius 
vector. 

Cor.  1. — The  general  integrals  of  (1)  will  contain  four 
arbitrary  constants.  One,  h,  that  was  introduced  in  (5), 
and  two  more  will  be  introduced  by  the  integration  of  (8). 
If  the  value  of  r  in  terms  of  Q,  deduced  from  the  integral 
of  (8),  be  substituted  in  (5),  and  that  equation  be  then 
integrated,  the  fourth  constant  will  be  introduced,  and  the 
path  of  the  particle  and  its  position  at  any  time  will  be 
obtained.  The  four  constants  must  be  determined  from 
the  initial  circumstances  of  motion ;  viz.,  the  initial 
position  of  the  particle,  depending  on  two  independent 
co-ordinates,  its  initial  velocity,  and  its  direction  of  pro- 
jection. 

Cor.  2. — By  means  of  (9)  we  may  ascertain  the  law  of 
the  force  which  must  act  upon  a  particle  to  cause  it  to 
describe  a  given  curve.  To  effect  this  we  must  determine 
the  relation  between  u  and  6  from  the  polar  equation  of  the 
orbit  referred  to  the  required  centre  as  pole  ;  we  must  then 
differentiate  u  twice  with  respect  to  6,  and  substitute  the 
result  in  the  expression  for  P,  eliminating  0,  if  it  occurs, 
by  means  of  the  relation  between  u  and  6.  In  this  way  wo 
shall  obtain  P  in  terms  of  u  alone,  and  therefore  of  t 
alone. 

Cor.  3. — When  we  know  the  relation  between  r  and  0 
from  (9),  we  may  by  (5)  determine  the  time  of  describing 
a  given  portion  of  the  orbit ;  or,  conversely,  find  the  posi« 
tion  of  the  particle  in  its  orbit  at  any  time.* 

*  See  Tait  and  Steele's  Dynamics  of  a  Particle,  p.  119;  also  Pratt's  Mech^ 
p.  222. 


THE  SECTIONAL   AREA.  325 

Cor.  4. — If  p  is  the  perpendicular  from  the  origin  to 
the  tangent  we  have  from  Calculus,  p.  176, 

xdy  —  y  dx  =  p  ds  ; 
which  in  (3)  gives 


(10) 


ds h9 

Jt  ~p' 

and  this  in  (6)  gives 

h2 

d-„  =  -2Pdr. 

Differentiating,  and  solving  for  P,  we  have 

^-^3     «fr>  (11) 

which  is  the  equation  of  the  orbit  betiveen  the  radius  vector 
and  the  perpendicular  on  the  tangent  at  any  point. 

182.  The  Sectorial  Area  Swept  over  by  the 
Radius  Vector  of  the  Particle  in  any  time  is  Pro- 
portional to  the  Time. — Let  A  denote  this  area ;  then  we 
have  from  Calculus,  p.  364, 

A  =  ifr*dd 

=  ifhdt,  by  (5)  of  Art.  181, 

if  A  and  t  be  both  measured  from  the  commencement  of 
the  motion.  Therefore  the  areas  swept  over  by  the  radius 
vector  in  different  times  are  proportional  to  the  times,  and 
equal  areas  trill  be  described  in  equal  times. 

Cor. — If  t  =s  1,   we  have  A  =  \h.    Hence  h  =  twice 
the  sectorial  area  described  in  one  unit  of  time. 

183.  The  Velocity  of  the   Particle  at  any  Point 
of  its  Orbit. — We  have  for  the  velocity, 


326  VELOCITY  AT  ANT  POINT  OF  THE   ORBIT. 

_  ds 
v  ~~  Tt 

=  -  by  (10)  of  Art.  181.         (I) 

Hence,  the  velocity  of  the  particle  at  each  point  of  its 
path  is  inversely  proportional  to  the  perpendicular  from  the 
centre  on  the  tangent  at  that  point. 

Cor.  1. — We  have,  by  Calculus,  p.  180, 

1        1       1^ 

^  -  r2  +  7A  d62 

o      du2    .  1 '   .   ,  „  rt„ . 

=       +  d&9  SmCe  T  =  u  (  ^ 

which  in  (1)  gives 

'-5 -*(-  +  »  ™ 

another  important  expression  for  the  velocity. 
Cor.  2.— From  (6)  of  Art.  181,  we  have 

d(^)  =  d(v*)  =  -2Pdr.  (3) 


Let  V  be  the  velocity  at  the  point  of  projection,  at 
which  let  r  =  R,  and  since  P  is  some  function  of  r,  let 
P  =f(r)}  then  integrating  (3)  we  get 

S= -».£>«*. 

.-.    ^-T»  =  2[/1(5)-/,(r)],  (4) 

which  is  another  expression  for  the  velocity  ;  and  since  this 
is  a  function  only  of  the  corresponding  distances,  R  and  r, 
it  follows   that  the  velocity  at   any  point  of  the  orbit  is 


VELOCITY  AT  ANY  POI.XT   OF  THE   ORBIT.  327 

independent  of  the  path  described,  and  depends  solety  on  the 
magnitude  of  the  attraction,  the  distance  of  the  point  from 
the  centre,  and  the  velocity  and  distance  of  projection. 

From  (4)  it  appears  that  the  velocity  is  the  same  at  all 
points  of  the  same  orbit  which  are  equally  distant  from  the 
centre;  if  r  =  R,  the  velocity  =  V;  and  thus  if  the  orbit 
is  a  re-entering  curve,  the  particle  always,  in  its  successive 
revolutions,  passes  through  the  same  point  with  the  same 
velocity. 

If  the  velocity  vanishes  at  a  distance  a  from  the  centre 
(4)  becomes 

■»  -  2  [,t\  (a)  -A  (r)]  (5) 

and  a  is  called  the  radius  of  the  circle  of  zero  velocity. 

Cor.  3.— From  (3)  we  have 

d  (v2)  =  -  2Pdr; 

.-.    vdv  =  —  Pdr.  (6) 

Taking  the  logarithm  of  (1)  we  have 

log  v  =  log  h  —  log  p. 

Differentiating  we  get 


dv 

V 

dp 

(?) 

)ividing  (6)  by  (7), 

we  get 

»*: 

=  2P 

p  dr 

'2dp 

==  2P  x  J  chord  of  curvature*  through 

the  centre ; 

(8) 

*  To  prove  that  §  —  is  one-fourth  the  chord  of  curvature. 
F  2  dp 

Let  MD  (Fig.  81),  he  the  tangent  to  the  orbit,  and  C  the  centre  of  curvature  ;  let 

OD  -p,  CM  =  p,  the  radius  of  curvature  ;  and  the  angle  MEN  =  <f>.    Then  MS,  the 


328  VELOCITY  AT  ANY  POINT   OF  THE   ORBIT. 

and,  comparing  this  with  (6)  of  Art.  140,  it  appears  that 
the  particle  at  any  point  has  the  same  velocity  which  it 
would  have  if  it  moved  from  rest  at  that  point  towards  the 
centre  of  force,  under  the  action  of  the  force  continuing 
constant,  through  one-fourth  of  the  chord  of  the  circle  of 
curvature. 

Hence,  the  velocity  of  a  particle  at  any  point  of  a  central 
orbit  is  the  same  as  that  ivhich  ivould  be  acquired  by  a 
particle  moving  freely  from  rest  through  one-fourth  of  the 
chord  of  curvature  at  that  point,  through  the  centre,  under 
the  action  of  a  constant  force  whose  magnitude  is  equal  to 
that  of  the  central  attraction  at  the  point. 

Cor.  4. — If  the  orbit  is  a  circle  having  the  centre  of  force 

part  of  the  radius  vector  OM,  which  is  intercepted  by  the  circle  of  curvature  is 
caLed  the  chord  of  curvature.    Its  value  is  determined  as  follows ; 


a> 

<2) 

to 

ft 


We  have  (Fig.  81) 

Prom  Calculus,  p.  180,  (10),  we 
and 

<t>  =  e  +  OMD 
=  0  +  sin-if  ; 

•.    d4>  =  d0  +  rdp-pdr. 
r  tfr*~—p* 
have 

de  =       ^r    -, 
r  Vra — pa 

fig  _  r°de  _       rdr 

Substituting  (9)  in  (1)  we  get 

df-       dp 

But  Calculus,  p.  221,  we  have 
p  = 

yra  —p2 

Now  MS  (Fig.  81)  =  2MC  sin  OMD, 


=  **-*$.*» 


the  chord  of  curvature ;  therefore 
p  dr 


.    =  one-fourth  the  chord  of  curvature. 
dp 


THE   ORBIT   UNDER    VARIABLE  ATTRACTION.        329 

in  the  centre,  and  R,  V,  P,  are  respectively  the  radius, 
velocity  and  central  force,  we  have 

V2  =  PR. 

Cor.  5.— From  (5)  of  Art.  181,  we  have 

dO       h 


dt 


(9) 


The  first  member,  being  the  actual  velocity  of  a  point 
on  the  radius  vector  at  the  unit's  distance  from  the  centre, 
is  the  angular  velocity  of  the  particle  (Art.  160).  Hence 
the  angular  velocity  of  a  particle  varies  inversely  as  the 
square  of  the  radius  vector. 

Sch. — A  point  in  a  central  orbit  at  which  the  radius 

vector  is  a  maximum  or  minimum  is  called  an  Apse ;  the 

radius  vector  at  an  apse  is  called  an  Apsidal  Distance  ;  and 

the  angle  between  two  consecutive  apsidal  distances  is  called 

an  Apsidal  Angle  of  the  orbit.     The  analytical  conditions 

du 
for  an  apse  are,  of  course,  that  -=-  =  0,  and  that  the  first 

uu 

derivative  which  does  not  vanish  should  be  of  an  even 
order.  The  first  condition  ensures  that  the  radius  vector 
at  an  apse  is  perpendicular  to  the  tangent. 

184.  The  Orbit  when  the  Attraction  Varies  In- 
versely as  the  Square  of  the  Distance. — A  particle  is 
projected  from  a  given  point  in  a  given  direction  with  a  given 
velocity,  and  moves  under  the  action  of  a  central  attraction 
varying  inversely  as  the  square  of  the  distance  ;  to  determine 
the  orbit. 

Let  the  centre  of  force  be  the  origin ;  V  =  the  velocity 
of  projection  ;  R  =  the  distance  of  the  point  of  projection 
from  the  origin;  j9  =  the  angle  between  R  and  the  line  of 


330        THE   ORBIT   UNDER    VARIABLE  ATTRACTION, 

projection ;   and  let  \i  =  the   absolute  force  and    t  =  fl 
when  the  particle  is  projected.     Then  since  the  velocity  = 

-  (Art.  183),  and  at  the  point  of  projection  p  =  R  sin  <3, 

we  have 

r=nLe>h=™sinl3'  w 

As  the  force  varies  inversely  as  the  square  of  the  distance, 
we  have 

P  =  ^2  =  ^,  (since  r  =  i).  (2) 

which  in  (9)  of  Art.  181  gives 

w  + u  =  w  ® 

Multiplying  by  2du  and  integrating,  we  get 

du2        9       _  \i 

_  +  «.  =  8^.  +  ,j 

1        1  du*  V2 

when  t  =  0,  u  =  -  =  -^,  and  -p  -f  u?  =  -jjf  (Art  183, 

Cor.  1) ;  therefore 

Substituting  this  value  for  c  we  get 

^  + u  -     h2B     +  h%  m 

Therefore  (Art.  183,  Cor.  1)  we  have 

(velocity)*  =  V*  +  fy  (*  -  ^)  (5) 


THE    ORBIT   UNDER    VARIABLE   ATTRACTION.        331 

which  shows  that  the  velocity  is  the  greatest  when  r  is  the 
least,  and  the  least  when  r  is  the  greatest. 
Changing  the  form  of  (4)  we  have 


du*     vm  -  2 


a+n-te-)'     « 


dd2  h2R 

To  express  this  in  a  simpler  form,  let 

jp  =  by  and  — ^> — -  +  ^  =  c2 ;  and  (6)  becomes 


=  d0, 


[C2  _  (W  _  J)2]i 

the  negative  sign  of  the  radical  being  taken.     Integrating 

we  have, 

--  *  —  &        a       J 

cos  * =  0  —  #, 

c 

where  c'  is  an  arbitrary  constant; 

.-.    u  =  b  +  ccos  (9  —  £r).  (7) 

Replacing  in  (7)  the  values  of  £  and  c,  and  the  value  of  h, 
from  (1),  and  dividing  both  terms  of  the  second  member  by 
(i,  we  have  for  the  equation  of  the  path, 

1  +  j  \  ( V2R  -  2p)  R  V2  sin*  0  -f  1  Tcos  (0—c') 


which  is  the  equation  of  a  conic  section,  the  pole  being  at 
the  focus,  and  the  angle  (6  —  c')  being  measured  from  the 


332        THE   ORBIT   UNDER    VARIABLE  ATTRACTION. 

shorter  length  of  the  axis  major.  For  if  e  is  the  eccentricity 
of  a  conic  section,  r  the  focal  radius  vector,  and  <f>  the 
angle  between  r  and  that  point  of  a  conic  section  which  is 
nearest  the  focus,  we  have, 

i  =  u  =  i +  ««■»,  (9) 

Comparing  (8)  and  (9),  we  see  that 

63  =  \(V*R  -  2//)  RV*sm*0  +  1;  (10) 

<p  =  d-c'.  (11) 

Now  the  conic  section  is  an  ellipse,  parabola,  or  hyper- 
bola, according  as  e  is  less  than,  equal  to,  or  greater  than 
unity;  and  from  (10)  e  is  less  than,  equal  to,  or  greater 
than,  unity  according  as  V2R  —  2/x  is  negative,  zero,  or 
positive ;  therefore  we  see  that  if 

2fi 
V2  <  ~,  e  <  1,  and  the  orbit  is  an  ellipse,         (12) 

2u 
V9  =  -£  e  =  1,  and  the  orbit  is  a  parabola,        (13) 

2u 

V2  >  -£,  e  >  1,  and  the  orbit  is  a  hyperbola.      (14) 

Cor.  1. — By  (1)  of  Art.  173,  we  see  that  the  square  of 
the  velocity  of  a  particle  falling  from  infinity  to  a  distance 
R  from  the  centre  of  force,  for  the  law  of  attraction  we 

2u 

are  considering,  is  -£•     Hence  the  above  conditions  may 
K 

be  expressed    more  concisely    by  saying  that    the  orbit, 

described  about  this  centre  of  force,  will  be  an  ellipse,  a 


THE   OliBIT  AN   ELLIPSE. 


333 


parabola,  or  a  hyperbola,  according  as  the  velocity  is  less 
chan,  equal  to,  or  greater  than,  the  velocity  from  infinity. 

The  species  of  conic  section,  therefore,  does  not  depend 
on  the  position  of  the  line  in  which  the  particle  is  pro- 
jected, but  on  the  velocity  of  projection  in  reference  to  the 
distance  of  the  point  of  projection  from  the  centre  of 
force. 


Cor.  2. — From  (11),  we  see  that  0  —  c'  is  the  angle 
between  the  focal  radius  vector, 
r,  and  that  part  of  the  principal 
axis  which  is  between  the  focus 
and  the  point  of  the  orbit  which 
is  nearest  to  the  focus ;  i.  e.,  it 
is  the  angle  PFA  (Fig.  82) ;  and 
therefore  if  the  principal  axis  is  the  initial  line  c'  =  0. 

185.   Suppose  the  Orbit  to  be  an  Ellipse. — Here 
V2  <  ^-;  so  that  from  (10)  we  have 


e2  =  1  —  -2  (2^  —  V*R)  RV*  sin* 0. 


(1) 


Now  the  equation  of  an  ellipse,  where  r  is  the  focal 
radius  vector,  0  the  angle  between  r  and  the  shorter  seg- 
ment of  the  major  axis,  2a  the  major  axis,  e  the  eccen- 
tricity, is 

a  (1  -  &)  d 


r  = 


1  +  e  cos  0 ' 


21 


+ 


cos  0 


a  (1  —  e2)  T  a  (1  —  e2) ' 
comparing  (2)  with  (8)  of  Art.  184,  we  have 


(8) 


a  (l  —  e2)  ~~  iPF'sitfiS' 


334  THE    ORBIT  AN  ELLIPSE. 

substituting  for  1  —  e2  its  value  from  (1),  and  solving  foi 
a,  we  have 

a  =  fyT^VW'  (3) 

which  shows  that  the  major  axis  is  independent  of  the  direc- 
tion of  projection. 

We  may  explain  the  several  quantities  which  we  have 
used,  by  Fig.  82. 

B  is  the  point  of  projection ;  FB  =  R ;  DB  is  the  line 
along  which  the  particle  is  projected  with  the  velocity  V; 
FBD  =  (3,  the  angle  of  projection ;  FP  =  r ;  PFA  =  d . 
FD  =  R  sin  (5 ;  if  (3  =  90°,  the  particle  is  projected  from 
an  apse,  i.  e.,  from  A  or  A'. 

Cor.  1. — To  determine  the  apsidal  distances,  FA  and 
FA',  we  must  put  ~  =  0,  (Art.  183,  Sch.),  and  (4)  of 
Art.  184  give  us  the  quadratic  equation 

U        h*U  +  WR       &  -°'  ]4) 

the  tivo  roots  of  which  are  the  reciprocals  of  the  tioo  apsidal 
distances,  a  (1  —  e)  and  a  (1  +  e). 

Cor.  2. — Since  the  coefficient  of  the  second  term  of  (4) 
is  the  sum  of  the  roots  with  their  signs  changed,  we  have 


1 

i 

1 

2fi 

a  (I  - 

-.•)   ' 

0(1  + 

■)" 

h*' 

•  • 

«(1  - 

-*)  = 

0 

V1 

(5) 

which  gives  the  latus  rectum  of  the  orliL 


KEPLER  'S  LA  WS.  335 

Cor.  3. — From  Art.  182  we  have,  calling  7*  the  time, 

rp  %A 

T 

where  A  is  the  area  swept  over  by  the  radius  vector  in  the 
time  T.  Therefore  for  the  time  of  describing  an  ellipse, 
we  have 

T  _  2  area  of  ellipse 


27ra«  Vl  -  e*  _ 

— =,  from  (5), 

VafM  (1  -  e2) 


V  t*      ■ 

which  is  the  time  occupied  by  the  particle  \n  passing  from 
any  point  of  the  ellipse  around  to  the  same  point  again.* 

186.  Kepler's  Laws. — By  laborious  calculation  from 
an  immense  series  of  observations  of  the  planets,  and  of 
Mars  in  particular,  Kepler  enunciated  the  following  as  the 
laws  of  the  planetary  motions  about  the  Sun. 

/.  Tlie  orbits  of  the  planets  are  ellipses,  of  which 
the  Sun  occupies  a  focus. 

II.  The  radius  vector  of  each  planet  describes 
equal  areas  in  equal  times. 

III.  The  squares  of  the  periodic  times  of  the 
planets  are  as  the  cubes  of  the  major  axes  of  their 
orbits. 

187.  To  Determine  the  Nature  of  the  Force  which 
Acts    upon    the    Planetary   System.— (1)    From   the 

*  Called  Periodic  Time. 


336  PLANETARY  SYSTEM. 

second  of  these  laws  it  follows  that  the  planets  are  retained 
in  their  orbits  by  an  attraction  tending  to  the  Sun. 

Let  (x,  y)  be  the  position  of  a  planet  at  the  time  t 
referred  to  two  co-ordinate  axes  drawn  through  the  Sun  in 
the  plane  of  motion  of  the  planet ;  X,  Y,  the  component 
accelerations  due  to  the  attraction  acting  on  it,  resolved 
parallel  to  the  axes ;  then  the  equations  of  motions  are 

^_  T.  <py  -  v 

df>  ~       '     dP  ~      ' 

•••  x§-yS  =  xY-yx-         (1) 

But,  by  Kepler's  second  law,  if  A  be  the  area  described 

dA 
by  the  radius  vector,  -j-  is  constant, 


dA 

"'•     dt 

or  ^^r 

""  *Vdt 

y  —  J  ==  a  constaot. 

ifferentiating,  we  have 

xd*y 
xdt* 

d2x 

.-.    xY- 

yX  =  0,  from 

d)> 

•  • 

X      x 
Y~y' 

which  shows  that  the  axial  components  of  the  acceleration, 
due  to  the  attraction  acting  on  the  planet,  are  proportional 
to  the  co-ordinates  of  the  planet;  and  therefore,  by  the 
parallelogram  of  forces  (Art.  30),  the  resultant  of  JTand  Y 
passes  through  the  origin. 


PLANETARY  SYSTEM.  337 

Hence  the  forces  acting  on  the  planets  all  pass  through 
the  Sim's  centre. 

(2)  From  the  first  of  these  laws  it  follows  that  the 
central  attraction  varies  inversely  as  the  square  of  the 
distance. 

The  polar  equation  of  an  ellipse,  referred  to  its  focus,  is 

_    a  (1  -  e2) 
~~  1  +  e  cos  d9 


or 


1  +  e  cos  0 
a  (1  -  e2)  ' 


Hence  -—  +  u  = 


dd*    '        . .  a  (1  -  e2)  ' 

and  therefore,  if  P  is  the  attraction  to  the  focus,  we  have 
[Art.  181,  (9)], 

A2         1 


a  (1  _  e2)  r5 

Hence,  i/  ^Ae  or&iY  fo  #w  ellipse,  described  about  a  centre 
of  attraction  at  the  focus,  the  law  of  intensity  is  that  of  the 
inverse  square  of  the  distance. 

(3)  From  the  third  law  it  follows  that  the  attraction  of 
the  Sun  (supposed  fixed)  which  acts  on  a  unit  of  mass  of 
each  of  the  planets,  is  the  same  for  each  planet  at  the  same 
distance. 

By  Art.  185,  Cor.  3,  we  have 

4tt2 
T2  =  —  a\ 
u 

15 


338  EXAMPLES. 

But  by  the  third  law,  T2  <x  az,  and  therefore  \i  must  be 
constant ;  i.  e.,  the  strength  of  attraction  of  the  Sun  must 
be  the  same  for  all  the  planets.  Hence,  not  only  is  the  law 
of  force  the  same  for  all  the  planets,  but  the  absolute  force 
is  the  same. 

This  very  brief  discussion  of  central  forces  is  all  that  we 
have  space  for.  To  pursue  these  enquiries  further  would 
compel  us  to  omit  matters  that  are  more  especially  entitled 
to  a  place  in  this  book.  The  student  who  wishes  to  pursue 
the  study  further  is  referred  to  Tait  and  Steele's  Dynamics 
of  a  Particle,  or  Price's  Anal.  Mech's,  Vol.  I,  or  to  any 
work  on  Mathematical  Astronomy.  We  shall  conclude 
with  the  following  examples. 

EXAMPLES. 

1.  A  particle  describes  an  ellipse  under  an  attraction 
always  directed  to  the  centre  ;  it  is  required  to  find  the  law 
of  the  attraction,  the  velocity  at  any  point  of  the  orbit,  and 
the  periodic  time. 

(1)  The  polar  equation  of  the  ellipse,  the  pole  at  the 
centre,  is 

2       cos**?       sin20 

?f-3r  +  nr;  (i) 

and     u  <m  +  m  =  \&  -  ?) (cos  6  - sm  e)-      ^ 

But  [Art.  181,  (9)]  we  have 


EXAMPLES.  339 

by  (3), 

(cos2  0  —  sin2  0)],  by  (2), 

by  factoring, 

and  therefore  the  attraction  varies  directly  as  the  distance. 
If  [i  =  the  absolute  force  we  have,  by  (4), 

P  ==  fi  a*b\  (5) 

(2)  If  v  =  the  velocity,  we  have,  by  Art.  183, 

7i2       h*bn 
*=^^(Anal.Geom.,p.l38) 

=  fib'%  by  (5), 
where  V  is  the  semi-diameter  conjugate  to  r. 
.  • .     v  =  b'  ^/fi. 

(3)  If  T  =  the  periodic  time,  we  have,  by  Art.  182, 

m       2nab        2tt  .  . 

2'=-T-  =  v?,by(5), 

and  hence  the  periodic  time  is  independent  of  the  magni- 
tude of  the  ellipse,  and  depends  only  on  the  absolute 
central  attraction.     (See  Tait  and  Steele's  Dynamics  of  a 


340  EXAMPLES. 

Particle,    p.     144,    also    Trice's    Anal.    Mech's,     \7ol.     I, 
p.  51G.) 

2.  A  particle  describes  an  ellipse  under  an  attraction 
always  directed  to  one  of  the  foci  ;  it  is  required  to  find  the 
law  of  attraction,  the  velocity,  and  the  periodic  time. 

(1)  Here  we  have 

__  1  -f  e  cos  0  du  _  e  sin  6    #        -  . 

U  -    a  (1  -  ea7  '     '""    dd  "  """a(l  — e*);        "' 

,  rf3w        —  ^  cos  0 

and  dW  "  «(1  -  ^* 

which  in  (9)  of  Art.  181  gives 

o  (1  -  e2)  ~~  a  (1  -  e2)  *  r2;  '  ' 

hence  the  attraction  varies  inversely  as  the  square  ot  the 
distance.     If  fi  =  the  absolute  force,  we  have  by  (2) 

7/2  =  fia  (1  -  e2).  (3J 

(2)  By  Art.  183,  Cor.  1,  we  have 

1  9   ,   du2        %au  —  1     u    /-v  ,^x 

&  =  u3  +  dm=zW(r=rF)>hjW>  <4> 

...    i.=|  =  t(!^:i),by(3)a11d(4).  (5) 

(3)  If  T  =  the  periodic  time  we  have  (Art.  182) 

r==  27^(1-^)* 


_  2*0?  (1  —  e2)^  __  _2_tt_    | 
"  \ua  (1  —  e2)"l*  ""  a/k 


(6) 


EXAMPLES.  341 

and  hence  the  periodic  time  varies  as  the  square  root  of 
the  cube  of  the  major  axis. 

3.  Find  the  attraction  by  which  a  particle  may  describe 
a  circle,  and  also  the  Telocity,  and  the  periodic  time,  (1) 
when  the  centre  of  attraction  is  in  the  centre  of  the  circle, 
and  (2)  when  the  centre  of  attraction  is  in  the  circum- 
ference. 

(1)  Let  a  =  the  radius;  then  the  polar  equation,  the 
pole  at  the  centre,  is 

1     du       $u 

Also  *  =  ?'    and     T=~?.  (2) 

From  (1)  and  (2)  we  have 

a 

and  hence  the  central  attraction  is  equal  to  the  square  of 
the  velocity  divided  by  the  radius  of  the  circle.* 

(2)  The  equation,  is 

r  =  2a  cos  0  ;     .  • .    2au  =  sec  6, 

gad  u  +  -^5  =  8  A3 : 

do* 

D  Q    27,2    B  8«2^ 

and  hence  the  attraction   varies  inversely    as    the    fifth 

*  Called  the  Centrifugal  Force.    See  Art.  198. 


342  EXAMPLES. 

power  of  the  distance  ;   and  if  \i  =  the  absolute  force,  we 
have  fi  =  Sa2h2; 

.  • .    h2  =  J~ ;    and    v2  =  ^~ 

8#2 '  2r* 

If  T  =  the  periodic  time,  we  have 

r  =  — ^-     (See  Price's  Anal.  Mech.,  Vol.  I.,  p.  518.) 
u2 

4.  Find  the  attraction  by  which  a  particle  may  describe 
the  lemniscate  of  Bernouilli  and  also  the  velocity,  and  the 
time  of  describing  one  loop,  the  centre  of  attraction  being 
in  the  centre  of  the  lemniscate,  and  the  equation  being 
r2  =  a2  cos  20. 

.         _        3/i2a*      2         ft      m       /3\*  , 
Ans.  P  =  ■ — r-  ;  v2  =  ■—--,  T  =    -)  aK 
r*  3r6  \/V 

5.  Find  the  attraction  by  which  a  particle  may  describe 
the  cardioid  and  also  the  velocity,  and  the  periodic  time, 
the  equation  being  r  =  a  (1  +  cos  6). 

3ah2,    2_2/z.    rp_  /3V\* 


Ans.  P  =  — t  ;  v2 


3V\* 

7T. 


ri    >        -  3rs>  \  M 

6.  Find  the  attraction  by  which  a  particle  may  describe 
a  parabola,  and  also  the  velocity,  the  centre  of  attraction 

2a 

being  at  the  focus,  and  the  equation  being  r  =  = v 

Ans.  P  =  ^  ;  v2  =  ^-     Compare  (18)  of  Art.  184. 

7.  Find  the  attraction  by  which  a  particle  may  describe 

a  hyperbola,  and  the  velocity,  the  centre  of  attraction  being 

a  (e2 \\ 

at  the  focus,  and  the  equation  being  r  =  - — —• 

1  &  1  -f  e  cos  0 

a  (1  —  e2)  r2  a 


EXAMPLES.  343 

8.  If  the  centre  of  attraction  is  at   the  centre  of   the 

hyperbola,  find  the  attraction,  and  velocity,  the  equation 

cos2  0       sin2  6 
being  -^ ^  =  n\ 

h2 
Ans.  P  = -^  r  =  —  |ur;  z;2  =  ^  (r2  —  a2  -f-  J2). 

9.  Find  the  attraction  to  the  pole  under  which  a  particle 

will  describe  (1)  the  curve  whose  equation  is  r  =  2a  cos  nd, 

2a 

and  (2)  the  curve  whose  equation  is  r  = • 

v  '  l  1  —  e  cos  nd 

^   (1)  P  =  -Ji"  +  S^S     W  *  =  <^T2  + 

(1  _  w2\  ^.2 

-£— —     That  is,  the  attraction  in  the  first  curve  varies 

partly  as  the  inverse  fifth  power,  and  partly  as  the  inverse 
cube,  of  the  distance  ;  and  in  the  second  it  varies  partly  as 
the  inverse  square,  and  partly  as  the  inverse  cube,  of  the 
distance. 

10.  A  planet  revolved  round  the  sun  in  an  orbit  with  a 
major  axis  four  times  that  of  the  earth's  orbit ;  determine 
the  periodic  time  of  the  planet.  Ans.  8  years. 

11.  If  a  satellite  revolved  round  the  earth  close  to  its 
surface,  determine  the  periodic  time  of  the  satellite. 

Ans. of  the  moon's  period. 

(60)* 

12.  A  body  describes  an  ellipse  under  the  action  of  a 
force  in  a  focus  :  compare  the  velocity  when  it  is  nearest 
the  focus  with  its  velocity  when  it  is  furthest  from  the 
focus. 

A ns.  As  1  -\-  e  :  1  —  e,  where  e  is  the  eccentricity. 

13.  A  body  describes  an  ellipse  under  the  action  of  a 
force  to  the  focus  8\  if  H  be  the  other  focus  show  that  the 


344  EXAMPLES. 

velocity  at  any  point  P  may  be  resolved  into  two  velocities, 
respectively  at  riglit  angles  to  SP  and  HP,  and  each  vary- 
ing its  HP. 

14.  A  body  describes  an  ellipse  under  the  action  of  a 
force  in  the  centre:  if  the  greatest  velocity  is  three  times 
the  least,  find  the  eccentricity  of  the  ellipse.   Ana.  §  V2. 

15.  A  body  describes  an  ellipse  under  the  action  of  a 
force  in  the  centre  :  if  the  major  axis  is  20  feet  and  the 
greatest  velocity  20  feet  per  second,  find  the  periodic  time. 

Ans.  77  seconds. 


16.  Find  the  attraction  to  the  pole  under  which  a  par- 

1 
r3' 


tide  may  describe  an  equiangular  spiral.         .         _.      1 


ii 

17.  If  P  =  ^  (5r2  —  8c2),   and  a  particle  be  projected 

from  an  apse  at  a  distance  c  with  the  velocity  from  infinity ; 
prove  that  the  equation  of  the  orbit  is 

r  =  \  (<P  +  e~#). 

18.  If  P  =  2[i  ( rJ,  and  the  particle  be  projected 

from  an  apse  at  a  distance  a  with  velocity ,  prove  that 

it  will  be  at  a  distance  r  after  a  time 


1    (^.log!liLSSE*4.rVir 


CHAPTER     III. 

CONSTRAINED     MOTION. 

188.  Definitions. — A  particle  is  constrained  in  its  mo- 
tion when  it  is  compelled  to  move  along  a  given  fixed  curve 
or  surface.  Thus  far  the  suhjects  of  motion  have  been 
particles  not  constrained  by  any  geometric  conditions,  but 
free  to  move  in  such  paths  as  are  due  to  the  action  of  the 
impressed  forces.  AVe  come  now  to  the  case  of  the  motion 
of  a  particle  which  is  constrained ;  that  is,  in  which  the 
motion  is  subject,  not  only  to  given  forces,  but  to  undeter- 
mined reactions.  Such  cases  occur  when  the  particle  is  in 
a  small  tube,  either  smooth  or  rough,  the  bore  oi  which  is 
supposed  to  be  of  the  same  size  as  the  particle ;  or  when  a 
small  ring  slides  on  a  curved  wire,  with  or  without  friction  ; 
or  when  a  particle  is  fastened  to  a  string,  or  moves  on  a 
given  surface.  If  we  substitute  for  the  curve  or  surface  a 
force  whose  intensity  and  direction  are  exactly  equal  to 
those  of  the  reaction  of  the  curve,  the  particle  will  describe 
the  same  path  as  before,  and  we  may  treat  the  problem  as 
if  the  particle  were  free  to  move  under  the  action  of  this 
system  of  forces,  and  therefore  apply  to  it  the  general  equa- 
tions of  motion  of  a  free  particle. 

189.  Kinetic  Energy  or  Vis  Viva  (Living  Force), 
and  Work. — A  particle  is  constrained  to  move  on  a  given 
smooth  plane  curve,  under  given  forces  in  the  plane  of  the 
curve,  to  determine  the  motion. 

Let  APC  be  the  curve  along  which  the  particle  is  com- 
pelled to  move  when  acted  upon  by  any  given  forces.  Let 
Ox  and  Oy  be  the  rectangular  axes  in   the  plane  of  the 


346 


KINETIC  ENER  G  1 '. 


curve,  the  axis  y  positive  up- 
wards, and  (x,  y)  the  place  of 
the  particle,  P,  at  the  time  t ; 
let  X,  Y,  parallel  respectively  to 
the  axes  of  x  and  y,  be  the  axial 
components  of  the  forces,  the 
mass  of  the  particle  being  m  ; 
let  R  be  the  pressure  between 
the  curve  and  particle,  which 
acts  in  the  normal  to  the  curve,  since  it  is  smooth.  Then 
the  equations  of  motion  are 


Fig. 83 


d2x        v       ndy 


(1) 


d&  ' 


£  =  Y  +  E 


dx 
ds' 


(3) 


Multiplying  (1)  and  (2)  respectively  by  dx  and  dy,  and 
adding,  we  have 


m- 


dx  d2x  -f-  dy  d2y 


dP 


=  Xdx  +  Ydy. 


Integrating  between  the  limits  t  and  tQf  and  calling  vQ  the 
initial  velocitv,  we  have 


*-%vt*'=f(Xdx+Tds) 


(3) 


iffi 
The  term  —  v2  is  called  the  vis  viva*,  or  Kinetic  Energy 

of  the  mass  m;  that  is,  vis  viva  or  kinetic  energy  is  a 
quantity  which  varies  as  the  product  of  the  mass  of  the 
particle  and  the  square  of  its  velocity.  There  is  particular 
advantage  in  defining  vis  viva,  or  kinetic  energy,  as  half 


*  See  Thomson  and  Tait's  Nat.  Phil.,  p.  222. 


KINETIC  ENERGY,  34? 

the  product  of  the  mass  and  the  square  of  its  velocity.* 
The  first  member,  therefore,  of  (3)  is  the  vis  viva  or  kinetic 
energy  of  m  acquired  in  its  motion  from  (x0,  y0)  to  (x,  y) 
under  the  action  of  the  given  forces. 

The  terms  Xdx  and  Ydy  are  the  products  of  the  axial 
components  of  the  forces  by  the  axial  displacements  of  the 
mass  in  the  time  clt,  and  are  therefore,  the  elements  of  work 
done  by  the  accelerating  forces  X  and  Y  in  the  time  dt, 
according  to  the  definition  of  work  given  in  Art.  101,  Rem.; 
so  that  the  second  member  of  (3)  expresses  the  work  done 
by  these  forces  through  the  spaces  over  which  they  moved 
the  mass  in  the  time  between  t0  and  t.  This  equation  is 
called  the  equation  of  kinetic  energy  and  of  work;  it  shows 
that  the  work  done  by  a  force  exerting  action  through  a 
given  distance,  is  equal  to  the  increase  of  kinetic  energy 
which  has  accrued  to  the  mass  in  its  motion  through  that 
distance. 

If  in  the  motion,  kinetic  energy  is  lost,  negative  work  is 
done  by  the  force  ;  i.  e.,  the  work  is  stored  up  as  potential 
work  in  the  mass  on  which  the  force  has  acted.  Thus,  if 
work  is  spent  on  winding  up  a  watch,  that  work  is  stored 
in  the  coiled  spring,  and  is  thus  potential  and  ready  to  be 
restored  under  adapted  circumstances-  Also,  if  a  weight  is 
raised  through  a  vertical  distance,  work  is  spent  in  raising 
it,  and  that  work  may  be  recovered  by  lowering  the  weight 
through  the  same  vertical  distance. 

This  theorem,  in  its  most  general  form,  is  the  modern 
principle  of  conservation  of  energy  ;  and  is  made  the  funda- 
mental theorem  of  abstract  dynamics  as  applied  to  natural 
philosophy. 

In  this  case  we  have  an  instance  of  space-integrals,  which, 
as  we  have  seen,  gives  us  kinetic  energy  and  work ;  the 
solution  of  problems  of  kinetic  energy  and  work  will  be 
explained  in  Chap.  V. 

*  Some  writers  define  vis  viva  as  the  whole  product  of  the  mass  and  the  square 
of  the  velocity.    See  Routh's  Rigid  Dynamics,  p.  259. 


348  REACTION  OF  THE   CONSTRAINING    CURVE. 

Now  if  X  and  Y  are  functions  of  the  co-ordinates  x  and 
y  the  second  member  of  (3)  can  be  integrated  ;  let  it  be  the 
differential  of  some  function  of  x  and  y,  as  </>  (x,  y).  Inte- 
grating (3)  on  this  hypothesis,  and  supposing  v  and  v0  to 
be  the  velocities  of  the  particle  at  the  points  (x,  y)  and 
(x0,  y0)  corresponding  to  t  and  t0,  we  have 


TYl 

-  (v*  -  v*)  =  <p  (x,  y)  -  0  (x0,  y0)  (4) 


which  shows  that  the  kinetic  energy  gained  by  the  particle 
constrained  to  move,  under  the  forces  X,  Y,  along 
any  path  whatever,  from  the  point  (a?0,  y0)  to  the  point 
(x,  y),  is  entirely  independent  of  the  path  pursued,  and 
depends  only  upon  the  co-ordinates  of  the  points  left  and 
arrived  at;  the  reaction  R  does  not  appear,  which  is  clearly 
as  it  should  be,  since  it  does  no  work,  because  it  acts  in  a 
line  perpendicular  to  the  direction  of  motion. 

190.  To  Find  the  Reaction  of  the  Constraining 
Curve. — For  convenience,  the  muss  of  the  particle  may  be 
taken  as  unity.     Multiplying  (1)  and  (2)  of  Art.  189  by 

fly  fix 

—  and  -=-,  subtracting  the  former  from  the  latter,  and 
solving  for  R}  we  have, 

„       cPy  dx  —  cPx  dy   ,    ^dy       ^dx 
dt2  as  as  as 

in  which  p  is  the  radius  of  curvature  at  the  point  P.  The 
last  two  terms  of  (1)  are  the  normal  components  of  the 
impressed  forces ;  and  therefore,  if  the  particle  were  at  rest, 
they  would  denote  the  whole  pressure  on  the  curve ;  but 


REACTION  OF  THE   CONSTRAINING    CURVE.  349 

the  particle  being  in  motion,  there  is  an  additional  pressure 

on  the  curve  expressed  by  — • 

In  the  above  reasoning  we  have  considered  the  particle  to 
be  on  the  concave  side  of  the  curve,  and  the  resultant  of  X 
and  l"to  act  towards  the  convex  side  along  some  line  as  PP 
so  as  to  produce  pressure  against  the  curve.  If  on  the 
contrary,  this  resultant  acts  towards  the  concave  side,  along 
PF'  for  example,  then,  whether  the  particle  be  on  the 
concave  or  convex  side,  the  pressure  against  the  curve  will 

be  the  difference  between  —  and  the  normal  resultant  of  X 

P 
and  Y. 

191.  To  Find  the  Point  where  the  Particle  Will 
Leave  the  Constraining  Curve.— It  is  evident  that  at 
that  point,  R  =  0,  as  there  will  be  no  pressure  against  the 
curve.     Therefore  (1)  of  Art.  190  becomes 

o   " "       els  ds 

=  F'  cos  F'PR 

if  F'  be  the  resultant  of  X  and  Y, 

.-.    v*  —  F'p  cos  F'PR 

=  2F'-  \  chord  of  curvature  in  the  direction  FF\ 

Comparing  this  with  (6)  of  Art.  140,  we  see  that  the 
particle  will  leave  the  curve  at  the  point  where  its  velocity  is 
such  as  would  be  produced  by  the  resultant  force  then  acting 
on  it,  if  continued  constant  during  its  fall  from  rest  through 
a  space  equal  to  \  of  the  chord  of  curvature  parallel  to  that 
resultant.  (See  Tait  and  Steele's  Dynamics  of  a  Particle, 
p.  170.) 


350  CONSTRAINED  MOTION. 

192.  Constrained  Motion  Under  the  Action  of 
Gravity. — When  gravity  is  the  only  force  acting  on  the 
particle,  the  formulae  are  simplified.  Taking  the  axis  of  y 
vertical  and  positive  downwards,  the  forces  become 

X  =  0,    and     Y  =  +  g; 

and  for  the  velocity  we  have,  by  (3)  of  Art.  189, 

¥2  -  iV  =  9  (V  —  y0)  (!) 

where  y0  is  the  initial  space  corresponding  to  the  time  t0. 
For  the  pressure  on  the  curve  we  have,  by  (1)  of  Art.  190, 

n  t?    .        dx 

If  the  origin  be  where  the  motion  of  the  particle  begins, 
the  initial  velocity  and  space  are  zero,  and  (1)  becomes 

I*  =  gy.  (3) 

This  shows  that  the  velocity  of  the  particle  at  any  time 
is  entirely  independent  of  the  form  of  the  curve  on  which 
£t  moves ;  and  depends  solely  on  the  perpendicular  distance 

?ougli  which  it  falls. 

193.  Motion  on  a  Circular  Arc  in  a  Vertical 
Plane. — Take  the  vertical  diameter  as  axis  of  y,  and  its 
lower  extremity  as  origin  ;  then  the  equation  of  the  circle  is 


(i) 


a?  = 

2ay 

-tf\ 

dx 
a  — 

y  " 

dy  = 

X 

ds 
a 

MOTION  ON  A    CIRCULAR  ARC. 


351 


Let  (l;  h)  be  the  point  K  where 
the  particle  starts  from  rest,  and  (x,  y) 
the  point  P  where  it  is  at  the  time  t. 
Then  the  particle  will  have  fallen 
through  the  height  HM  =  h  —  y, 
and  hence  from  (3)  of  Art.  192  we 
iiave 

els 


( 

c 

\ 

i 

W 

Kv 

\  7 

ssvJ/l 

-Yp 

o 

X 

Fig. 

84   4 

dt 


v  =  Vty  (h  —  y). 


(2) 


Hence  the  velocity  is  a  minimum  when  y  =  h,  and  a 
maximum  when  y  =  0;  and  this  maximum  velocity  will 
carry  the  particle  through  0  to  K'  at  the  distance  h  above 
the  horizontal  line  through  0. 

To  find  the  time  occupied  by  the  particle  in  its  descent 
from  K  to  the  lowest  point,  0,  we  have  from  (2) 


dt  =  — 


ds 


V2g  (h  -  y) 


ady 


V%g(h  -  y)  (2ay-  y*) 


by(i)     (3) 


the  negative  sign  being  taken  since  t  is  a  decreasing  func- 
tion of  s. 

This  expression  does  not  admit  of  integration  ;  it  may  be 
reduced  to  an  elliptic  integral  of  the  first  kind,  and  tables 
are  given  of  the  approximate  values  of  the  integral  for 
given  values  of?/.* 

If,  however,  the  radius  of  the  circle  is  large,  and  the 
greatest  distance  KO,  over  which  the  particle  moves,  is 
small,  we  may  develope  (3)  into  a  series  of  terms  in  ascend- 

ing  powers  of  ^-,  and  thus  find  the  integral  approximately. 


*  See  Legendre's  Traite  des  Fonctions  Elliptiques. 


352 


THE   SIMPLE  PENDULUM, 


Let  The  the  time  of  motion  of  the  particle  from  iT to  K\ 
i.  e.,  from  y  =  h,  through  y  =  0,  to  y  =  h  again,  then  (3) 
becomes 


V  0  Jh  </hy  —  y2\ 


1\ 

2a/ 


-i 


dy 


Vhy  -  y2 


integrating  each  term  separately  we  have 


+ 


which  is  the  complete  expression  for  the  time  of  moving 
from  the  extreme  position  K  on  one  side  of  the  vertical  to 
the  extreme  position  K'  on  the  other;  this  is  called  an 
oscillation.     (See  Price's  Anal.  Mechs.,  Vol.  L,  p.  518.) 
If  the  arc  is  very  small,  h  is  very  small  in  comparison 

with  a,  and  all  the  terms  containing  —  will  be  very  small, 

and  by  neglecting  them  (4)  becomes 


—VI- 


(5) 


194.  The  Simple  Pendulum. — Instead  of  supposing 
the  particle  to  move  on  a  curve,  we  may  imagine  it  sus- 
pended by  a  string  of  invariable  length,  or  a  thin  rod 
considered  of  no  weight,  and  moving  in  a  vertical  plane 
about  the  point  C ;  for,  whether  the  force  acting  on  the 
particle  be  the  reaction  of  the  curve  or  the  tension  of  the 
string,  its  intensity  is  the  same,  while  its  direction,  in 
either  ea*e  i.s  along  the  normal  to  the  curve. 


RELATION  OF  TIME.   LENGTH,   ETC.  353 

When  the  particle  is  supposed  to  be  suspended  by  a 
thread  without  weight,  it  becomes  what  is  termed  a  simple 
pendulum;  and  although  such  an  instrument  can  never  be 
perfectly  attained,  but  exists  only  in  theory,  yet  approxima- 
tions may  be  made  to  it  sufficiently  near  for  practical  pur- 
poses, and  by  means  of  Dynamics  we  may  reduce  the 
calculation  of  the  motion  of  such  a  pendulum  to  that  of 
the  simple  pendulum. 

If  I  is  the  length  of  the  rod,  the  time  of  an  oscillation  is 
approximately  given  by  the  formula 


=  Wl 


9 


when  the  angle  of  oscillation  is  very  small,  *.  e.,  not  ex- 
ceeding about  4° ;  *  and  therefore,  for  all  angles  between 
this  and  zero,  the  times  of  oscillation  of  the  same  pen- 
dulum will  not  perceptibly  differ ;  i.  e.,  in  very  small  arcs 
the  oscillations  may  be  regarded  as  isochronal,  or  as  all 
performed  in  the  same  time. 

195.  Relation  of  Time,  Length,  and  Force  of 
Gravity. — From  (1)  of  Art.  194,  we  have  T'oc  y7if#is 

constant ;  Toe  — =  if  I  is  constant;  g  oc  I  if  Tis  constant, 

vg 

that  is 

(1)  For  the  same  place  the  times  of  oscillation  are  as  the 
square  roots  of  the  lengths  of  the  pendulums. 

(2)  For  the  same  pendulum  the  times  of  oscillation  are 
inversely  as  the  square  roots  of  the  force  of  gravity  at 
different  places. 

*  If  the  initial  inclination  is  5°,  the  second  term  of  (4)  is  only  0.000476  ;  if  1°  the 
second  term  is  only  0.000019. 


354  HEIGHT  OF  MOUNTAIN  DETERMINED. 

(3)  For  the  same  time  the  lengths  of  pendulums  vary  as 
the  force  of  gravity. 

Hence  by  means  of  the  pendulum  the  force  of  gravity  at 
different  places  of  the  earth's  surface  may  be  determined. 
Let  L  be  the  length  of  a  pendulum  which  vibrates  seconds 
at  the  place  where  the  value  of  g  is  to  be  found  ;  then  from 
(1)  of  Art.  194  we  have 


Vf; 


•-.    g  =  n*L;  (1) 


and  from  this  formula  g  has  been  calculated  at  many  places 
on  the  earth.  The  method  of  determining  L  accurately 
will  be  investigated  in  Chap.  VII. 

Cor. — If  n  be  the  number  of  vibrations  performed  dur- 
ing N  seconds,  and  T  the  time  of  one  vibration, 

then  n  =  ^,  by  (1)  of  Art.  194  =  ^x/j'        (2) 

Since  gravity  decreases  according  to  a  known  law,  as  we 
ascend  above  the  earth's  surface,  the  comparison  of  the 
times  of  vibration  of  the  same  pendulum  on  the  top  of  a 
mountain  and  at  its  base,  would  give  approximately  its 
height. 

196.  The  Height  of  a  Mountain  Determined  with 
the  Pendulum. — A  seconds  pendulum  is  carried  to  the  top 
of  a  mountain  ;  required  to  find  the  height  of  the  mountain 
by  observing  the  change  in  the  time  of  oscillation. 

Let  r  be  the  radius  of  the  earth  considered  spherical;  h 
the  height  of  the  mountain  above  the  surface;  I  the  length 
of  the  pendulum  ;  g  and  g'  the  values  of  gravity  on  the 
earth's  surface,  and  at  the  top  of  the  mountain  respectively. 
Then  (Art.  174)  we  have 


HEIGHT  OF  MOUNTAIN   DETERmNED.  355 

g'  ~  \     r     /'     •  '    9   -  (r  +  h)*>  {l) 

which  is  the  force  of  gravity  at  the  top  of  the  mountain. 

Let  n  =  the  number  of  oscillations  which  the  Beconda 
pendulum  at  the  top  of  the  mountain  makes  in  24  hours; 

«       «     «         *        -n  r  24  x  00  x  60      TJ  „ 

then  the  time  ot  oscillation  = Hence  trom 

n     . 

(1)  of  Art.  195,  we  have 


• 


24  x  60  x  60  I  r  +  h        I         /1X 

=  n\/  -  =  w — ! —  \/-,  by  (1); 

n  \  g'  r     V  9     J  w 

A        24  x  60  x  60  /  .  /F  \  ,_. 

...     -  = - -_  1,  (since  *y-  =  l),  (3) 

which  gives  the  height  of  the  mountain  in  terms  of  the 
radius  of  the  earth.  For  the  sake  of  an  exam'ple,  suppose 
the  pendulum  to  lose  5  seconds  in  a  day  ;  that  is.  to  make 
5  oscillations  less  than  it  would  make  on  the  surface  of  the 
earth. 

Then  n  =  24  x  60  x  60  —  5 ; 

which  in  (2)  gives 

h  24  x  60  x  60 


r        24  x  60  x  60 


1 


-  I1  ~  24  x  60  x  "12/    ~  X  ~  24  x  60  x  12  "early ; 


4000  .       ..  . 

-  =  £  mile,  nearly, 


24  x  60  x  12 

;  oeing  4000  miles  (approximately). 

197.  The  Depth  of  a  Mine  Determined  by  Ob- 
serving the  Change  of  Oscillation  in  a  Seconds 
Pendulum. — Let  r  be  the  radius  of  the  earth  as  in  the 


356  CENTRIPETAL   FORCE. 

last  case ;  h  the  depth  of  the  mine  ;  g  and  g'  the  values  of 
gravity  on  the  earth's  surface  and  at  the  bottom  of  the 
mine.     Then  (Art.  171)  we  have 

g'  -  r-h  V* 

Let  n  =  the  number  of  oscillations  which  the  seconds 
pendulum  at  the  bottom  of  the  mine  makes  in  24  hours. 

„.                   24  x  60  x  60* 
Then 


V. 


g(r-  h) 


r 

—  < 

r  —  h 


h 


h  _    /  n  Y 

r  "  V24  x  60  x  60/ ' 


from  which  h  can  be  found.    If,  as  before,  the  pendulum 
loses  5  seconds  a  day,  we  have 

h 


S-H 


r  \        24  x  60  x  12> 


-  nearly, 


12  x  60  x  12 

.  • .    h  =  \  mile  nearly. 

(See  Price's  Anal.  Mech's,  Vol.  I,  p.  590,  also  Pratt's 
Mech's,  p.  376.) 

198.  Centripetal   and   Centrifugal   Forces. — Since 

v2 
the  pressure  — ,  at  any  point,  depends  entirely  upon  the 

velocity  at  that  point  and  the  radius  of  curvature,  it  would 
remain  the  same  if  the  forces  X  and  Y  were  both  zero,  in 
which   case   it   would   be   the   whole  normal  pressure,  R, 


CENTRIFUGAL   FORCE,  357 

against  the  curve.      It  is  easily  seen,  therefore,  that  this 

pressure  arises  entirely  from  the  inertia  of   the    moving 

particle,  i.  e.,  from  its  tendency  at  any  point,  to  move  in 

the  direction  of  a  tangent ;  and  this  tendency  to  motion 

along  the  tangent  necessarily  causes  it  to  exert  a  pressure 

against  the  deflecting  curve,  and  which  requires  the  curve 

v2 
to  oppose  the  resistance  —  •    Hence,  since  the  particle  if 

Left  to  itself,  or  if  left  to  the  action  of  a  force  along  the  tan- 
gent, would,  by  the  law  of  inertia,  continue  to  move  along 

v2 
that  tangent,  —  is  the  effect  of  the  force  which  deflects  the 

particle  from  its  otherwise  rectilinear  path,  and  draws  it 
towards  the  centre  of  curvature.  This  force  is  called  the 
Centripetal  Force,  which,  therefore,  may  be  defined  to  be 
the  force  ivhich  deflects  a  particle  from  its  otherwise  recti' 
linear  path.  The  equal  and  opposite  reaction  exerted  away 
from  the  centre  is  called  the  Centrifugal  Force,  which  may 
be  defined  to  be  the  resistance  ivhich  the  inertia  of  a  particle 
in  motion  opposes  to  whatever  deflects  it  from  its  rectilinear 
path.  Centripetal  and  centrifugal  are  therefore  the  same 
quantity  under  different  aspects.  The  action  of  the  former 
is  towards  the  centre  of  curvature,  while  that  of  the  latter 
is  from  the  centre  of  curvature.  The  two  are  called  central 
forces.  They  determine  the  direction  of  motion  of  the  par- 
ticle but  do  not  affect  the  velocity,  since  they  act  continu- 
ally at  right  angles  to  its  path.  If  a  particle,  attached  to  a 
string,  be  whirled  about  a  centre,  the  intensity  of  these 
central  forces  is  measured  by  the  tension  of  the  string.  If 
the  string  be  cut,  the  particle  will  move  along  a  tangent  to 
the  curve  with  unchanged  velocity. 

Coe.   1. — If  m  be  the  mass  moving  with  velocity  v,  ita 

v2 
centrifugal  force  is  m  —      If   w  be  the  angular  velocity 


358  CENTRIFUGAL   FORCE. 

described  by  the  radius  of  curvature,  then  (Art.  1G0,  Ex.  1), 

v  =  ow,  and  consequently 

the  centrifugal  force  of  m  =  mcfip.  (1) 

Cor.  2. — Let  m  move  in  a  circle  with  a  constant  velocity, 
v ;  let  a  =  the  radius  of  the  circle,  and  T  the  time  of  a 
complete  revolution;  then  2na  =  vT; 

4  77% 
.*.    the  centrifugal  force  of  m  =  m  -=■ ;  (2) 

and  thus  the  centrifugal  force  in  a  circle  varies  directly  as 
the  radius  of  the  circle,  and  inversely  as  the  square  of  the 
periodic  time. 

Cor.  3. — If  m  moves  in  the  circle  with  a  constant 
angular  velocity,  w,  then  (Art.  160,  Ex.  1),  v  =  ad) ; 

.*.    the  centrifugal  force  of  m  =  muPa;  (3) 

and  therefore  varies  directly  as  the  radius  of  the  circle. 

Thus  if  a  particle  of  mass  m  is  fastened  by  a  string  of 
length  a  to  a  point  in  a  horizontal  plane,  and  describes  a 
circle  in  the  plane  about  the  given  point  as  centre,  the  cen- 
trifugal force  produces  a  tension  of  the  string,  and  if  w  is 
the  constant  angular  velocity,  the  tension  =  m  (*)2a. 

199.  The  Centrifugal  Force  at  the  Equator. — Let 

U  denote  the  equatorial  radius  of  the  earth  =  20926202* 
feet,  T  the  time  of  revolution  upon  its  axis  =  86164 
seconds,  and  n  =  3.1415926.  Substituting  these  values  in 
(2)  of  Art.  198,  and  denoting  the  centrifugal  force  at  the 
equator  by/,  and  the  mass  by  unity,  we  have 

/=—-  =0.11186  feet.  .  (1) 

*  Ency.  Brit.,  Art.  Geodesy. 


CENTR I  FUG  A  L   FOR  CE. 


359 


The  force  of  gravity  at  the  equator  lias  been  found  to  be 
32.09022;  if  this  force  were  not  diminished  by  the  cen- 
trifugal force ;  i.  e.,  if  the  earth  did  not  revolve  on  its 
ads  the  force  of  gravity  at  the  equator  would  be 


G  =  32.09022  +  0.11120  =  32.20148  feet. 


(*) 


To  determine  the  relation  between  the  centrifugal  force 
and  the  force  of  gravity,  we  divide  (1)  by  (2)  which  gives 


/  _    ^11126^        J^ 

289 


•0148  ~  555*  nearly> 


(3) 


that  is,  the  centrifugal  force  at  the  equator  is  F|F  of  that 
which  the  force  of  gravity  at  the  equator  would  be  if  the 
ear tli  did  not  rotate. 


200.  Centrifugal  Force  at  Differ- 
ent Latitudes  on  the  Earth. — Let 

P  be  any  particle  on  the  earth's  surface 
describing  a  circumference  about  the 
axis,  NS,  with  the  radius  PD.  Let 
<p  =z  ACT  =  the  latitude  of  P;  R 
the  radius,  AC,  of  the  earth  ;  and  R' 
the  radius  PD  of  the  parallel  of  lati- 
tude passing  through  P.     Then  we  have 

R'  =  R  cos  0. 


/          D 

M 

\              C 

/A' 

s . 
Fig.  85 


(1) 


Let  the  centrifugal  force  at  the  point  P,  which  is  exerted 
in  the  direction  of  the  radius  DP,  be  represented  by  the 
line  PB.  Resolve  this  into  the  two  components  PF,  act- 
ing along  the  tangent,  and  PE,  acting  along  the  normaL 
Then  by  (2)  of  Art.  198  we  have 


PB 


^R' 

4:TT2R   COS   0 
2*2 


,  by  (1). 


(3) 


360  CENTRIFUGAL   FORCE. 

Hence,  the  centrifugal  force  at  any  point  on  the  earth's 
surface  varies  directly  as  the  cosine  of  the  latitude  of  the 
place. 

For  the  normal  component  we  have 

PE  =  PB  cos  0 

4:n2R  cos2  0  .      /n. 

=  — yir-  -  by  (3) 

=  /cos2  0,  by  (1)  of  Art.  199.    (3) 

Hence,  the  component  of  the  centrifugal  force  which  directly 
opposes  the  force  of  gravity,  at  any  point  on  the  earth's  sur- 
face, is  equal  to  the  centrifugal  force  at  the  equator,  mul- 
tiplied by  the  square  of  the  cosine  of  the  latitude  of  the 
place. 

Also  PF  =  PB  sin  0 

4=n2R  sin  0  cos  0   .      ,._. 
t        = jv- >  by  (2) 

=  |  sin  20,  by  (l)  oflrt.  199  ;  (4) 

that  is,  the  component  of  the  centrifugal  force  which  tends 
to  draw  particles  from  any  parallel  of  latitude,  P,  towards 
the  equator,  and  to  cause  the  earth  to  assume  the  figure 
of  an  oblate  spheroid,  varies  as  the  sine  of  twice  the 
latitude. 

The  preceding  calculation  is  made  on  the  hypothesis  that 
the  earth  is  a  perfect  sphere,  whereas  it  is  an  oblate 
spheroid  ;  and  the  attraction  of  the  earth  on  particles  at 
its  surface  decreases  as  we  pass  from  the  ^yoles  to  the 
equator.      The    pendulum    furnishes   the    most    accurate 


THE    CONICAL    rENDULUM. 


3G1 


method   of  determining   the  force  of  gravity  at  diffcren! 
places  on  the  earth's  surface. 

201.  The  Conical  Pendu- 
lum.— The  Governor. — Suppose 
a  particle,  P,  of  mass  m,  to  be  at- 
tached to  one  end  of  a  string  of 
length  I,  the  other  end  of  which  is 
fixed  at  A.  The  particle  is  made 
to  describe  a  horizontal  circle  of 
radius  PO,  with  uniform  velocity 
round  the  vertical  axis  AO,  so  that 
it  makes  u  revolutions  per  second. 
It  is  required  to  find  the  inclina- 
tion, d,  of  the  string  to  the  vertical, 
and  the  tension  of  the  string. 

The  velocity  of  P  in  feet  per  second  =  2rra  ■  OP  =  2nn  I 
sin  6.  The  forces  acting  upon  it  are  the  tension,  T,  of  the 
string,  the  weight,  m,  of  the  particle,  and  the  centrifugal 

force,  m =- -. — - —  (Art.  198).     Henca  resolving,  we  have 

for  horizontal  .forces,     T  sin  6  =  m  •  4tn2n2 1  sin  0  ;     (1) 
for  vertical  forces,         T  cos  0  =  mg.  (2) 


Fig. 86 


From  (1)  T  =  m-lirWl, 

which  in  (2)  gives 


cos  6  = 


9 


4tt2^  V 


(3) 
(4) 


where  Tand  6  are  completely  determined. 

If  the  string  be  replaced  by  a  rigid  rod,  which  can  turn 
about  A  in  a  ball  and  socket  joint,  the  instrument  is  called 
a  conical  pendulum,  and   occurs  in   the  governor  of  the 
steam-engine. 
16 


3G2  EXAMPLES. 


EXAMPLES 


1.  If  the  length  of  the  seconds  pendulum  be  39.1393 
inches  in  London,  find  the  value  of  g  to  three  places  of 
decimals.  Arts.  32.191  feet. 

2.  In  what  time  will  a  pendulum  vibrate  whose  length  is 
15  inches  ?  Ans.  0.62  sec.  nearly. 

3.  In  what  time  will  a  pendulum  vibrate,  whose  length  is 
double  that  of  a  seconds  pendulum  ?  Ans.  1.41  sees. 

4.  How  many  vibrations  will  a  pendulum  3  feet  long 
make  in  a  minute?  Ans.  62.55. 

5.  A  pendulum  wrhich  beats  seconds,  is  taken  to  the  top 
of  a  mountain  one  mile  high  ;  it  is  required  to  find  the 
number  of  seconds  which  it  will  lose  in  12  hours,  allowing 
the  radius  of  the  earth  to  be  4000  miles.    Ans.  10.8  sees. 

6.  What  is  the  length  of  a  pendulum  to  beat  seconds  at 
the  place  where  a  body  falls  16-j^  ft.  in  the  first  second  ? 

Ans.  39.11  ins.  nearly. 

7.  If  39.11  ins.  be  taken  as  the  length  of  the  seconds 
pendulum,  how  long  must  a  pendulum  be  to  beat  10  times 
in  a  minute  ?  Ans.  117^-  ft. 

8.  A  particle  slides  down  the  arc  of  a  circle  to  the 
lowest  point ;  find  the  velocity  at  the  lowest  point,  if  the 
angle  described  round  the  centre  is  60°.  Ans.   Vgr. 

9.  A  pendulum  which  oscillates  in  a  second  at  one  place, 
is  carried  to  another  place  where  it  makes  120  more  oscil- 
lations in  a  day ;  compare  the  force  of  gravity  at  the  latter 
place  with  that  at  the  former.  A?is.   (ffff)2. 

10.  Find  the  number  of  vibrations,  w1? which  a  pendulum 
will  gain  in  JV  seconds  by  shortening  the  length  of  the 
pendulum. 


EXAMPLES.  363 

Let  the  length,  I,  be  decreased  by  a  small  quantity, 
/,.  and  let  n  be  increased  by  nx  ;  then  from  (2)  of  Art.  105 

we  get 

n  rir 

»  +  *i  =  ^Vrr775 

which,  divided  by  (2)  of  Art.  195,  gives 

nr1  =  \t=tj  =  V-t)  =l  +  ii nearly- 

Hence  nt  =  —j» 

11.  If  a  pendulum  be  45  inches  long,  how  many  vibra- 
tions will  it  gain  in  one  day  if  the  bob*  be  screwed  up  one 
turn,  the  screw  having  32  threads  to  the  inch  ? 

Ans.  28. 

12.  If  a  clock  loses  two  minutes  a  day,  how  many  turns 
to  the  right  hand  must  we  give  the  nut  in  order  to  correct 
its  error,  supposing  the  screw  to  have  50  threads  to  the 
inch?  Ans.   5-4  turns. 

13.  A  mean  solar  day  contains  24  hours,  3  minutes, 
56-5  seconds,  sidereal  time  ;  calculate  the  length  of  the 
pendulum  of  a  clock  beating  sidereal  seconds  in  London. 
See  Ex.  1.  Ans.  38-925  inches. 

14.  A  heavy  ball,  suspended  by  a  fine  wire,  vibrates  in  a 
small  arc  ;  48  vibrations  are  counted  in  3  minutes.  Cal- 
culate the  length  of  the  wire.  Ans.  45-87  feet. 

15.  The  height  of  the  cupola  of  St.  Paul's,  above  the 
floor,  is  340  ft.;  calculate  the  number  of  vibrations  a  heavy 
body  would  make  in  half  an  hour,  if  suspended  from  the 
dome  by  a  fine  wire  which  reaches  to  within  G  inches  of 
the  floor.  Ans.  176-4. 

*  The  lower  extremity  of  the  pendulum. 


3G4  EXAMPLES. 

16.  A  seconds  pendulum  is  carried  to  the  top  of  a 
mountain  m  miles  high  ;  assuming  that  the  force  of 
gravity  varies  inversely  as  the  square  of  the  distance  from 
the  centre  of  the  earth,  find  the  time  of  an  oscillation. 

/4000  4-ffli 
4000" 


Ans.   (—7777^-      I  sees. 


17.  Prove  that  the  lengths  of  pendulums  vibrating  dur- 
ing the  same  time  at  the  same  place  are  inversely  as  the 
squares  of  the  number  of  oscillations. 

18.  In  a  series  of  experiments  made  at  Harton  coal-pit,  a 
pendulum  which  beat  seconds  at  the  surface,  gained  2-J- 
beats  in  a  day  at  a  depth  of  1260  ft. ;  if  g  and  g'  be  the 
force  of  gravity  at  the  surface  and  at  the  depth  mentioned, 
show  that 

ff'  —  ff  _       i 

a         —   TS200* 

19.  A  pendulum  is  found  to  make  640  vibrations  at  the 
equator  in  the  same  time  that  it  makes  641  at  Greenwich  ; 
if  a  string  hanging  vertically  can  just  sustain  80  lbs.  at 
Greenwich,  how  many  lbs.  can  the  same  string  sustain  at 
the  equator?  Arts.  80 J  lbs.  about. 

20.  Find  the  time  of  descent  of  a  particle  down  the  arc 

of  a  cycloid,  the  axis  of  the  cycloid  being  vertical  and  vertex 

downward  ;  and  show  that  the  time  of  descent  to  the  lowest 

point  is  the  same  whatever  point  of  the  curve  the  particle 

starts  from.  /f 

Ans.  TT\/~- 
V  9 

21.  If  in  Ex.  20  the  particle  begins  to  move  from  the 
extremity  of  the  base  of  the  cycloid  find  the  pressure  at  the 
lowest  point  of  the  curve. 

Ans.  2g ;  i.  e.,  the  pressure  is  twice  the  wreight  of  the 
particle. 


EXAMPLES.  365 

22.  Find  the  pressure  on  the  lowest  point  of  the  curve 
in  Art.  193,  (1)  when  the  particle  starts  from  rest  at  the 
highest  point,  A,  (Fig.  84),  (2)  when  it  starts  from  rest  at 
the  point  B. 

Ans.  (1)  5g ;  (2)  Sg ;  i.e.,  (1)  the  pressure  is  five  times 
the  weight  of  the  particle  and  (2)  it  is  three  times  the 
weight  of  the  particle. 

23.  In  the  simple  pendulum  find  the  point  at  which  the 
tension  on  the  string  is  the  same  as  when  the  particle 
hangs  at  rest. 

Ans.  y  =  \li,  where  h  is  the  height  from  which  the 
pendulum  has  fallen. 

24.  If  a  particle  be  compelled  to  move  in  a  circle  with  a 
velocity  of  300  yards  per  minute,  the  radius  of  the  circle 
being  16  ft.,  find  the  centrifugal  force. 

Ans.  14»  06  ft.  per  sec. 

25.  If  a  body,  weighing  17  tons,  move  on  the  circum- 
ference of  a  circle,  whose  radius  is  1110  ft.,  with  a  velocity 
of  16  ft.  per  sec,  find  the  centrifugal  force  in  tons  (take 
g  =  32-1948).  A?is.  0-1217  ton. 

26.  If  a  body,  weighing  1000  lbs.,  be  constrained  to  move 
in  a  circle,  whose  radius  is  100  ft.,  by  means  of  a  string 
capable  of  sustaining  a  strain  not  exceeding  450  lbs.,  find 
the  velocity  at  the  moment  the  string  breaks. 

Ans.  38-06  ft.  per  sec. 

27.  If  a  railway  carriage,  weighing  7-21  tons,  moving  at 
the  rate  of  30  miles  per  hour,  describe  a  portion  of  a  circle 
whose  radius  is  460  yards,  find  its  centrifugal  force  in  tons. 

Ans.  0-314  ton. 

28.  If  the  centrifugal  force,  in  a  circle  of  100  ft.  radius, 
be  146  ft.  per  sec,  find  the  periodic  time. 

Ans.  5-2  sees. 


3GG  EXAMPLES. 

29.  If  the  centrifugal  force  be  131  ozs.,  and  the  radius 
of  the  circle  100  ft.,  the  periodic  time  being  one  hour,  find 
the  weight  of  the  body.  Am.  386-309  tons. 

30.  Find  the  force  towards  the  centre  required  to  make 
a  body  move  uniformly  in  a  circle  whose  radius  is  5  ft.: 
with  such  a  velocity  as  to  complete  a  revolution  in  5  sees. 

Ans.  — -• 
o 

31.  A  stone  of  one  lb.  weight  is  whirled  round  horizon- 
tally by  a  string  two  yards  long  having  one  end  fixed  ;  find 
the  time  of  revolution  when  the  tension  of  the  string  is  3  lbs. 

Ans.  2tt  \  /  -  sees. 

V  g 

32.  A  weight,  w,  is  placed  on  a  horizontal  bar,  OA, 
which  is  made  to  revolve  round  a  vertical  axis  at  0,  with 
the  angular  velocity  o);  it  is  required  to  determine  the 
position,  A,  of  the  weight,  when  it  is  upon  the  point  of 
sliding,  the  coefficient  of  friction  being  /. 

Ans.  OA  =z% 

33.  Find  the  diminution  of  gravity  at  the  Sun's  equator 
caused  by  the  centrifugal  force,  the  radius  of  the  Sun  being 
441000  miles,  and  the  time  of  revolution  on  his  axis  being 
607  h.  48  m.  Ans.  0-  0192  ft.  per  sec. 

34.  Find  the  centrifugal  force  at  the  equator  of  Mercury, 
the  radius  being  i570  miles,  and  the  time  of  revolutior 
24  h.  5  m.  Ans.  0-0435  ft.  per  sec. 

35.  Find  the  centrifugal  force  at  the  equator,  (1)  of 
Venus,  radius  being  3900  miles  and  time  of  revolution 
23  h.  21  m.,  (2)  of  Mars,  radius  being  2050  miles  and 
periodic  time  24  h.  37  m.,  (3)  of  Jupiter,  radius  being 
43500  miles  and  periodic  time  9  h.  56  m.,  and  (4)  of  Saturn 
radius  being  39580  miles  and  periodic  time  10  h.  29  m. 


EXAMPLES,  367 

Ans.  (1)  0-11504  ft.  per  sec;  (2)  0-05-44  ft.  per  sec; 
(3)  7-0907  ft.  per  sec;  (4)  5-  7924  ft.  per  sec 

3G.  Find  the  effect  of  centrifugal  force  in  diminishing 
gravity  in  the  latitude  of  00°.     [See  (3)  of  Art.  200). 

Ans.  0-  028  ft,  per  sec. 

37.  Find  (1)  the  diminution  of  gravity  caused  by  cen- 
trifugal force,  and  (2)  the  component  which  urges  particles 
towards  the  equator,  at  the  latitude  of  23°. 

Ans.  (1)  0-  09  ft.  per  sec;  (2)  0-04  ft.  per  sec 

38.  A  railway  carriage,  weighing  12  tons,  is  moving 
along  a  circle  of  radius  720  yards,  at  the  rate  of  32  miles 
an  hour;  find  the  horizontal  pressure  on  the  rails. 

Ans.  0-38  ton,  nearly. 

39.  A  railway  train  is  going  smoothly  along  a  curve  of 
500  yards  radius  at  the  rate  of  30  miles  an  hour ;  find  at 
what  angle  a  plumb-line  hanging  in  one  of  the  carriages 
will  be  inclined  to  the  vertical.  Ans.  2°  18'  nearly. 

40.  The  attractive  force  of  a  mountain  horizontally  is/ 
and  the  force  of  gravity  is#;  show  that  the  time  of  vibra- 

tion  of  a  pendulum  will  be  n\/  — — ;  a  being  the  length 

of  the  pendulum. 

41.  In  motion  of  a  particle  down  a  cycloid  prove  that  the 
vertical  velocity  is  greatest  when  it  has  completed  half  its 
vertical  descent. 

42.  When  a  particle  falls  from  the  highest  to  the  lowest 
point  of  a  cycloid  show  that  it  describes  half  the  path  in 
two-thirds  of  the  time. 

43.  A  railway  train  is  moving  smoothly  along  a  curve  at 
the  rate  of  60  miles  an  hour,  and  in  one  of  the  carriages  a 
pendulum,  which  would  ordinarily  oscillate  seconds,  is 
observed  to  oscillate  121  times  in  two  minutes.  Show  that 
the  radius  of  the  curve  is  very  nearly  a  quarter  of  a  mile. 


368  EXAMPLES. 

44.  One  end  of  a  string  is  fixed  ;  to  the  other  end  a 
particle  is  attached  which  describes  a  horizontal  circle  with 
uniform  velocity  so  that  the  string  is  always  inclined  at  an 
angle  of  G0C  to  the  vertical;  show  that  the  velocity  of  the 
particle  is  that  which  would  be  acquired  in  falling  freely 
from  rest  through  a  space  equal  to  three-fourths  of  the 
length  of  the  string. 

45.  The  horizontal  attraction  of  a  mountain  on  a  particle 
At  a  certain  place  is  such  as  would  produce  in  it  an  accelera- 
tion denoted  by  -•     Show  that  a  seconds  pendulum  at  that 

jalace  will  gain  — ^—  beats  in  a  day,  very  nearly. 

46.  In  Art.  201,  suppose  I  equal  to  2  ft.  and  m  to  be  20 
lbs.,  and  that  the  system  makes  10  revolutions  per  sec,  and 
0,=  32;  find  0  and  T. 

Ans.  6  =  cos-1  — 2;  T  =  500tt2  pounds. 

47.  A  tube,  bent  into  the  form  of  a  plane  curve,  revolves 
with  a  given  angular  velocity,  about  its  vertical  axis ;  it  is 
required  to  determine  the  form  of  the  tube,  when  a  heavy 
particle  placed  in  it  remains  at  rest  in  all  parts  of  the 
tube. 

(Take  the  vertical  axis  for  the  axis  of  y,  and  the  axis  of  x 
horizontal,  and  let  w  =  the  constant  angular  velocity). 
Ans.  x2u>2  =  2gy,  if  x  =  0  when  y  =  0,  i.  e.,  the  curve 
is  a  parabola  whose  axis  is  vertical  and  vertex  downwards. 

48.  A  particle  moves  in  a  smooth  straight  tube  which 
revolves  with  constant  angular  velocity  round  a  vertical 
axis  to  which  it  is  perpendicular,  to  determine  the  curve 
traced  by  the  particle. 

Let  to  =  the  constant  angular  velocity ;  aud  (r,  0)  the 
position  of  the  narfcinte  at  the  time  t.  ana  let  r  =  a  when 


EXAMPLES.  369 

/  =  0.      Then   since   the   motion   of    the  particle  is  due 
entirely  to  the  centrifugal  force,  we  have 

—  %=  G>2r  ;   —  ■=.  gj2  (V2  —  d?\ 
dP  '  dP  K  ' 

dv 
if  -jj  =  0,  when  r  =  «.     Hence  we  have 
at 


-*~* 


« 


CHAPTER     IV 

IMPACT. 

202.  An  Impulsive  Force.— Hitherto  we  have  con- 
sidered force  only  as  continuous,  i.  e.,  as  acting  through  a 
definite  and  finite  portion  of  time,  and  producing  a  finite 
change  of  velocity  in  that  time.  Such  a  force  is  measured 
at  any  instant  by  the  mass  on  which  it  acts  multiplied  by 
the  acceleration  which  it  causes.  If  a  particle  of  mass  in  be 
moving  with  a  velocity  v,  and  be  retarded  by  a  constant 
force  which  brings  it  to  rest  in  the  time  t,  then  the  measure 

YtlV 

of  this  force  is  —  (Art.  20).  Now  suppose  the  time  t  dur- 
ing which  the  particle  is  brought  to  rest  to  be  made  very 
small ;  then  the  force  required  to  bring  it  to  rest  must  be 
very  large ;  and  if  we  suppose  t  so  small  that  we  are  unable 
to  measure  it,  then  the  force  becomes  so  great  that  we  are 
unable  to  obtain  its  measure.  A  typical  case  is  the  blow  of 
a  hammer.  Here  the  time  during  which  there  is  contact  is 
apparently  infinitesimal,  certainly  too  small  to  be  measured 
by  any  ordinary  methods;  yet  the  effect  produced  is  con- 
siderable. Similarly  when  a  cricket  ball  is  driven  back  by 
a  blow  from  a  bat,  the  original  velocity  of  the  ball  is 
destroyed  and  a  new  velocity  generated.  Also  when  a  bul- 
let is  discharged  from  a  gun,  a  large  velocity  is  generated 
in  an  extremely  brief  time.  Forces  acting  in  this  way  are 
called  impulsive  forces.  An  impulsive  force  may  therefore 
he  defined  to  he  a  force  tuhich  produces  a  finite  change  of 
motion  in  an  indefinitely  hrief  time.  An  Impulse  is  the 
effect  of  a  hlow. 

In  such  cases  as  these  it  is  impossible  accurately  to 
determine   the   force   and   time;    but   we   can    determine 


IMPACT   OR    COLLISION.  371 

their  product,  or  Ft,  since  this  is  merely  the  change 
in  velocity  caused  by  the  blow  (Art.  20).  Hence,  in 
the  case  of  blows,  or  impulsive  forces,  we  do  not  attempt 
to  measure  the  force  and  the  time  of  action  separately,  but 
simply  take  the  whole  momentum  produced  or  destroyed,  as 
the  measure  of  the  impulse.  Because  impulsive  forces  pro- 
duce their  effects  in  an  indefinitely  short  time  they  are 
sometimes  called  instantaneous  forces,  i.  e.,  forces  requiring 
no  time  for  their  auction.  But  no  such  force  exists  in 
nature ;  every  force  requires  time  for  its  action.  There  is 
no  case  in  nature  in  which  a  finite  change  of  motion  is 
produced  in  an  infinitesimal  of  time  ;  for,  whenever  a 
finite  velocity  is  generated  or  destroyed,  a  finite  time  is 
occupied  in  the  process,  though  we  may  be  unable  to 
measure  it,  even  approximately. 

203.  Impact  or  Collision. — When  two  bodies  in  rela- 
tive motion  come  into  contact  with  each  other,  an  impact 
or  collision  is  said  to  take  place,  and  pressure  begins  to  act 
between  them  to  prevent  any  of  their  parts  from  jointly 
occupying  the  same  space.  This  force  increases  from  zero, 
when  the  collision  begins,  up  to  a  very  large  magnitude  at 
the  instant  of  greatest  compression.  If,  as  is  always  the 
case  in  nature,  each  body  possesses  some  degree  of  elasticity, 
and  if  they  are  not  kept  together  after  the  impact  by 
cohesion  or  by  some  artificial  means,  the  mutual  pressure 
between  them,  after  reaching  a  maximum,  will  gradually 
diminish  to  zero.  Tiie  whole  process  would  occupy  not 
greatly  more  or  less  than  an  hour  if  the  bodies  were  of  such 
dimensions  as  the  earth,  and  such  degrees  of  rigidity  as 
copper,  steel,  or  glass.  In  the  case,  however,  of  globes  of 
these  substances  not  exceeding  a  yard  in  diameter,  the 
whole  process  is  probably  finished  within  a  thousandth  of 
a  second.* 

The  impulsive  forces  are  so  much  more  intense  than  the 

*  Thomson  and  Tait's  Nat.  Phil.,  o.  274. 


372  DIRECT  AND    CENTRAL   IMPACT. 

ordinary  forces,  that  during  the  brief  time  in  which  the 
former  act,  an  ordinary  force  does  not  produce  an  effect 
comparable  in  amount  with  that  produced  by  an  impulsive 
force.  For  example,  an  impulsive  force  might  generate  a 
velocity  of  1000  in  less  time  than  one- tenth  of  a  second, 
while  gravity  in  one-tenth  of  a  second  would  generate  a 
velocity  of  about  three.  Hence,  in  dealing  with  the  eifects 
of  impulses,  finite  forces  need  not  be  considered. 

204.  Direct  and  Central  Impact. — When  two  bodies 
impinge  on  each  other,  so  that  their  centres  before  impact 
are  moving  in  the  same  straight  line,  and  the  common  tan- 
gent at  the  point  of  contact  is  perpendicular  to  the  line  of 
motion,  the  impact  is  said  to  be  direct  and  central.  When 
these  conditions  are  not  fulfilled,  the  impact  is  said  to  be 


When  two  bodies  impinge  directly,  one  upon  the  other, 
the  mutual  action  between  them,  at  any  instant,  must  be 
in  the  line  joining  their  centres ;  and  by  the  third  law 
(Art.  166),  it  must  be  equal  in  amount  on  the  two  bodies. 
Hence,  by  Law  II,  they  must  experience  equal  changes  of 
motion  in  contrary  directions. 

We  may  consider  the  impact  as  consisting  of  two  parts ; 
during  the  first  part  the  bodies  are  coming  into  closer  con- 
tact with  each  other,  mutually  displacing  the  particles  in 
the  vicinity  of  the  point  of  contact,  producing  a  compres- 
sion and  distortion  about  that  point,  which  increases  till  it 
reaches  a  maximum,  when  the  molecular  reactions,  thus 
called  into  play,  are  sufficient  to  resist  further  compression 
and  distortion.  At  this  instant  it  is  evident  that  the 
points  in  contact  are  moving  with  the  same  velocity.  No 
body  in  nature  is  perfectly  inelastic;  and  hence,  at  the 
instant  of  greatest  compression,  the  elastic  forces  of  resti- 
tntion  are  brought  into  action  ;  and  during  the  second  part 
of  the  impact  the  mutual  pressure,  produced  by  the  elastic 
forces,  which  were  brought  into  action  by  the  compression 


ELASTICITY   OF  BODIES.  373 

during  the  first  part  of  the  impact,  tend  to  separate  the 
two  bodies,  and  to  restore  them  to  their  original  form. 

205.  Elasticity  of  Bodies. — Coefficient  of  Resti- 
tution.— It  appears  from  experiment  that  bodies  may  be 
20m  pressed  in  various  degrees,  and  recover  more  or  less 
their  original  forms  after  the  compressing  force  has  ceased, 
this  property  is  termed  elasticity.  The  force  urging  th& 
approach  of  bodies  is  called  the  force  of  compression  ;  the 
force  causing  the  bodies  to  separate  again  is  called  the 
force  of  restitution.  Elastic  bodies  are  such  as  regain  a 
part  or  all  of  their  original  form  when  the  compressing 
force  is  removed.  The  ratio  of  the  force  of  restitution  to 
that  of  compression  is  called  the  Coefficient  of  Restitution.* 
It  has  been  found  that  this  ratio,  in  the  same  bodies,  is 
constant  whatever  may  be  their  velocities. 

When  this  ratio  is  unity  the  two  forces  are  equal,  and  the 
body  is  said  to  be  perfectly  elastic  ;  when  the  ratio  is  zero, 
or  the  force  of  restitution  is  nothing,  the  body  is  said  to  be 
non-elastic ;  when  the  ratio  is  greater  than  zero  and  less 
than  unity,  the  body  is  said  to  be  imperfectly  elastic.  There 
are  no  bodies  either  perfectly  elastic  or  perfectly  non-elas- 
tic, all  being  more  or  less  elastic. 

In  the  cases  discussed  the  bodies  will  be  supposed  spher- 
ical, and  in  the  case  of  direct  impact  of  smooth  spheres  it 
is  evident  that  they  may  be  considered  as  particles,  since 
they  are  symmetrical  with  respect  to  the  line  joining  their 
centres. 

The  theory  of  the  impact  of  bodies  is  chiefly  due  to 
Newton,  who  found,  in  his  experiments,  that,  provided  the 
impact  is  not  so  violent  as  to  make  any  sensible  indentation 
in  either  body,  the  relative  velocity  of  separation  after  the 
impact  bears  a  ratio  to  the  relative  velocity  of  approach 
before  the  impact,  which  is  constant   for   the   same   two 

*  Sometimes  called  Coefficient  of  Elasticity.    Todhunter's  Mech.,  p.  272. 


371  DIRECT  IMPACT   OF  INELASTIC  BODIES. 

bodies.  In  Newton's  experiments,  however,  the  two  bodies 
seem  always  to  have  been  formed  of  the  same  sub- 
stance. He  found  that  the  value  of  this  ratio  (the  coeffi- 
cient of  restitution),  for  balls  of  compressed  wool  was  about 
|,  steel  about  the  same,  cork  a  little  less,  ivory  f ,  glass  ||. 
The  results  of  more  recent  experiments,  made  by  Mr. 
Hodgkinson,  and  recorded  in  the  Report  of  the  British 
Association  for  1834,  show  that  the  theory  may  be  received 
as  satisfactory,  with  the  exception  that  the  value  of  the 
ratio;  instead  of  being  quite  constant,  diminishes  when  the 
velocities  are  very  large. 

206.  Direct  Impact  of  Inelastic  Bodies. — A  sphere 
of  mass  M,  moving  with  a  velocity  v,  overtakes  and  impinges 
directly  on  another  sphere  of  mass  M',  moving  in  the  same 
direction  with  velocity  v',  and  at  the  instant  of  greatest 
mutual  compression  the  spheres  are  moving  with  a  common 
velocity  V.  Determine  the  motion  after  impact,  and  the 
impulse  during  the  compression. 

Let  R  denote  the  impulse  during  the  compression,  which 
acts  on  each  body  in  opposite  directions  ;  and  let  us  sup- 
pose the  bodies  to  be  moving  from  left  to  right.  Then, 
since  the  impulse  is  measured  by  the  amount  of  momentum 
gained  by  one  of  the  impinging  bodies  or  lost  by  the  other 
(Art.  202),  we  have 

Momentum  lost  by  M  —  M  (v  —  V)  =  R,  (1) 

"       gained  by  M'  = M'  (V -  v')  =  R,        (2) 

...    M{v-  V)  =  31'  (V-v').  (3) 


Solving  (3)  for  V  we  get 
F  = 

which  in  (1)  or  (2)  gives 


_  Mv  +  M'v'  , 

y  —     M  +  M '  *  v  } 


DIRECT  IMPACT  OF  INELASTIC  BODIES,  375 

M  +  M  v  ' 

Hence  the  common  velocities  of  the  tivo  bodies  after  impact 
is  equal  to  the  algebraic  sum  of  their  momenta,  divided  by 
the  sum  of  their  masses,  and  also,  from  (4),  the  whole 
momentum  after  impact  is  equal  to  the  sum  of  the  momenta 
before. 

Cor.  1. — Had  the  balls  been  moving  in  opposite  direc- 
tions, for  example  had  M'  been  moving  from  right  to  left, 
v'  would  have  been  negative,  in  which  case  we  would  have 

_       Mv  -  M'v'  ,      p       MM'  (v  +  v')         ,.. 

V  =  ~^r? tit-  ;     and     R  =  — ^ — in— '•        (6) 

M  ■+-  M'  ■  M  +  M'  v  ' 

From  the  first  of  these  it  follows  that  both  balls  will  be 
reduced  to  rest  if 

Mv  =  M'v'; 

that  is,  if  before  impact  they  have  equal  and  opposite 
momenta. 

Cor.  2. — If  M'  is  at  rest  before  impact,  v'  =  0,  and  (4) 
becomes 

V  -  M  +  Jlf'  *'' 

If  the  masses  are  equal  we  have  from  (4)  and  (6) 

_        tf  -f  v'  v  —  v'  /nX 

r=^->  or  -%-'.  (8) 

according  as  they  move  in  the  same  or  in  opposite  direc- 
tions. 

207.  Direct  Impact  of  Elastic  Bodies.— When  the 

balls  are  elastic  the  problem  is  the  same,  up  to  the  instant 
of  greatest  compression,  as  if  they  were  inelastic ;  but  at 


376  DIRECT  IMPACT   OF  INELASTIC  BODIES. 

this  instant,  the  force  of  restitution,  or  that  tendency  which 
elastic  bodies  have  to  regain  their  original  form,  begins  to 
throw  one  ball  forward  with  the  same  momentum  that  it 
throws  the  other  back,  and  this  mutual  pressure  is  propor- 
tional to  R  (Art.  205). 

Let  e  be  the  coefficient  of  restitution  ;  then  during  the 
second  part  of  the  impact,  an  impulse,  eR,  acts  on  each 
ball  in  the  same  direction  respectively  as  R  acted  during 
the  compression.  Let  vx  and  v{  be  the  velocities  of  the 
balls  M  and  M'  when  they  are  finally  separated.  Then  we 
have,  as  before, 

Momentum  lost  by  M  =  M(V—  vx)  =  eR,       (1) 

gained  by  M'  =  M '  (v,;  —  V)  =  eR.     (2) 

From  (1)  we  have 
ir      eR 

Mv  +  M V  _       eM'         _ 
""    M  +  M'         M+  M'{V      V) 

by  (4)  and  (5)  of  Art.  206, 
M' 

=  v  ~  JT+TF ^  +  *)  &  "v^  (3) 

Similarly  from  (2)  we  have 

<  =  *'+  Mf~M,{1  +  e){V~V,)'  (4) 

which  are  the  velocities  of  the  balls  when  finally  separated. 

These  results  may  be  more  easily  obtained  by  the  con- 
sideration that  the  whole  impulse  is  (1  +  e)  R ;  for  this 
gives  at  once  the  whole  momentum  lost  by  M  or  gained  by 
M'  during  compression  and  restitution  as  follows : 

M(v-vx)  =  (1  +  e)R,  (5) 


DIRECT  IMPACT   OF  INELASTIC  BODIES.  37? 

and  M'  («/  —  v')  =  (1  +  e)  R.  (6) 

Substituting  in  (5)  and  (6)  the  Value  of  R  from  (5)  of  Art. 
206,  we  have  the  values  of  v  and  i\'  immediately. 

Cor.  1. — If  the  balls  are  moving  in  opposite  directions, 
v'  becomes  negative.  If  the  balls  are  non-elastic,  e  —  0, 
and  (3)  and  (4)  reduce  to  (4)  of  Art.  206,  as  they  should. 

Cor.  2. — If  the  balls  are  perfectly  elastic,  e  =  1,  and  (3) 
and  (4)  become 

Cor.  3. — Subtracting  (4)  from  (3)  and  reducing,  we  get 

vx  —  Vi   =  v  —  v'  —  (1  -f  e)  (z;  — -  v')t 

=  —e(v-  v').  (9) 

Hence,  the  relative  velocity  after  impact  is  — e  times  the 
relative  velocity  before  impact. 

Cor.  4.— Multiplying  (3)  and  (4)  by  M  and  M ',  respect- 
ively, and  adding,  we  get 

Mvx  +  M%'  =  Mv  +  M'v'.  (10) 

Hence,  as  in  Art.  206,  the  algebraic  sum  of  the  momenta 
after  impact  is  the  same  as  before;  i.e.,  there  is  no  mo- 
mentum lost,  which  of  course  is  a  direct  consequence  of  the 
third  law  of  motion  (Art.  169). 

Cor.  5. — Suppose  v'  =  0,  so  that  the  body  of  mass  if, 
moving  with  velocity  v,  impinges  on  a  body  of  mass  M '  at 
rest,  then  (3)  and  (4)  become 


378  LOSS   OF  KINETIC  ENERGY. 

?'i  =  -it i^  v>    and     v,   =  -^ ^f  v.       (11) 

Hence  the  body  which  is  struck  goes  onwards;  and  the 
striking  body  goes  onwards,  or  stops,  or  goes  backwards, 
according  as  M  is  greater  than,  equal  to,  or  less  than  eM'. 
If  M'  =  eM,  then  (11)  becomes 

Vi  =r  (1  —  e)  v,    and    #/  r=  v.  (12) 

Cor.  6.— If  M  =  M  and  e=l;  that  is,  if  the  balls 
are  of  equal  mass,  and  perfectly  elastic,*  then  (7)  and  (8) 
become,  respectively, 

vy  =  v'9    and    vt'  =  v;  (13) 

that  is,  the  balls  interchange  their  velocities,  and  the 
motion  is  the  same  as  if  they  had  passed  through  one 
another  without  exerting  any  mutual  action  whatever. 

Cor.  7. — If  M'  be  infinite,  and  v'  =  0,  we  have  the  case 
of  a  ball  impinging  directly  upon  a  fixed  surface ;  substi- 
tuting these  values  in  (3)  it  becomes 

vx  =  —  ev ;  (14) 

that  is,  the  Ml  rebounds  from  the  fixed  surface  with  a  veloc- 
ity e  times  that  with  which  it  impinged. 

208.  Loss  of  Kinetic  Energy  f  in  the  Impact  of 
Bodies. — Squaring  (9)  of  Art.  207,  and  multiplying  it  by 
MM' ,  we  have 

MM'  [vx  -  vx'f  =  MM'  &{y  —  v'f 

=  MM'  (v  -  v'f  -  (1  -  e2)  MM'  (v  -  v'f.        (1) 

*  This  is  the  usual  phraseology,  hut  misleading,  Ency.  Brit.,  Vol.  XV,  Art 
Mech's. 

t  See  Art.  189. 


LOSS   OF  KINETIC  ENERGY.  379 

Squaring' (10)  of  Art.  207,  we  have 

(Mv,  +  M'vx')2  =  (Mv  +  M'v'f.  (2) 

Adding  (1)  and  (2),  we  get 

(M  +  M')  (Mv<>  +  JfV»j  =  (M  +■  Jf ')  (Mv2  +  Jf  V») 

-  (1  -  e2)  MM'  (v  -  t;')« ; 
.-.    \Mv?  +  pfV  =  \Mv2  4  ii/V*      ' 

JfJf' 

-i(l-*2)^^(*-^(3) 

the  last  term  of  which  is  the  loss  of  kinetic  energy  by 
impact,  since  e  can  never  be  greater  than  unity.  Hence, 
there  is  always  a  loss  of  kinetic  energy  by  impact,  except 
when  e  =  1,  in  which  case  the  loss  is  zero ;  i.  e.,  when  the 
coefficient  of  restitution  is  unity,  no  kinetic  energy  is  lost. 
When  e  =  0  the  loss  is  the  greatest,  and  equal  to 

\     MM'     , 

iwTM'^-^y  w 

From  (3)  we  see  that  during  compression  kinetic  energy 

MM' 

to  the  amount  of  i  ^ ^.—,  (v  —  v')2  is  lost ;    and  then 

2  M  +  M' v  ' 

during  restitution,  e2  times  this  amount  is  regained. 

Rem. — From  the  theory  of  kinetic  energy  it  appears 
that,  in  every  case  in  which  energy  is  lost  by  resistance, 
heat  is  generated ;  and  from  Joule's*  investigations  we 
learn  that  the  quantity  of  heat  so  generated  is  a  perfectly 
definite  equivalent  for  the  energy  lost ;  and  also  that,  in 

*  See  "The  Correlation  and  Conservation  of  Forces,"  by  Helmholtz,  Faraday, 
Liebig,  etc.  ;  also  M  Heat  as  a  Mode  of  Motion,"  by  Prof.  Tyndall.  Also  B.  Stewart's 
"  Conservation  of  Energy." 


380  OBLIQUE  IMPACT. 

any  natural  action,  there  is  never  a  development  of  energy 
which  cannot  be  accounted  for  by  the  disappearance  of  an 
equal  amount  elsewhere  by  means  of  some  known  physical 
agency.  Hence,  the  kinetic  energy  which  appears  to  be 
lost  in  the  above  cases  of  impact,  is  only  transformed, 
partly  into  heating  the  bodies  and  the  surrounding  air,  and 
partly  into  sonorous  vibrations,  as  in  the  impact  of  a  ham- 
mer on  a  bell. 

209.  Oblique  Impact  of  Bodies.— The  only  other 
case  which  we  shall  treat  of  is  that  of  oblique  impact  when 
the  bodies  are  spherical  and  perfectly  smooth. 

A  particle  impinges  with  a  given  velocity,  and  in  a  given 
direction,  on  a  smooth  plane;  required  to  determine  the 
motion  after  impact. 

Let  AC  represent  the  direc- 
tion of  the  velocity  before  im- 
pact, meeting  the  plane  at  C, 

and    CB    the    direction   after 

impact.      Draw   CD   perpen-  Fig. 87 

dicular    to   the   plane ;    then 

since  the  plane  is  smooth  its  impulsive  reaction  will  be 
along  CD. 

Let  v  and  vx  denote  the  velocities  before  and  after 
impact,  respectively ;  and  let  a  and  (3  denote  the  angles 
ACD  and  BCD. 

Resolve  v  along  the  plane  and  perpendicular  to  it.  The 
former  will  not  be  altered,  since  the  impulsive  force  acts 
perpendicular  to  the  plane  ;  the  latter  may  be  treated  as  in 
the  case  of  direct  impact,  and  will  therefore,  after  impact, 
be  e  times  what  it  was  before  (Art.  207,  Cor.  7).  Hence, 
resolving  vx  along,  and  perpendicular  to  the  plane,  we 
have 

vx  sin  (3  =  v  sin  a,  (1) 

v,  cos  f3  =  —  e  v  cos  a.  (2) 


OBLIQUE  IMPACT.  381 

Dividing  (2)  by  (1),  we  get 

cot  J3  =  —  e  cot  a.  (3) 

Squaring  (] )  and  (2),  and  adding,  we  get 

vx2  =  v2  (sin2  a  -f  e2  cos2  «).  (4) 

Thus  (3)  determines  the  direction,  and  (4)  the  magnitude 
of  the  velocity  after  impact. 

The  angle  ACD  is  called  the  angle  of  incidence,  and  the 
angle  BCD  the  angle  of  reflexion. 

Cor.  1. — If  the  elasticity  be  perfect,  or  e  =  1,  we  have 
from  (3)  and  (4), 

cot  /?  =  —  cot  a,  or  J3  =  —a',  (5) 

and  v?  =  v2,  or  vx  =  v.  (6) 

Hence,  in  perfectly  elastic  balls  the  angles  of  incidence 
arid  reflexion  are  numerically  equal,  and  the  velocities  before 
and  after  impact  are  equal.  This  is  the  ordinary  rule  in 
the  case  of  a  billiard  ball  striking  the  cushion. 

Cor.  2.— Suppose  e  —  0;  then  from  (3),  (3  =  90°. 
Thus,  if  there  is  no  elasticity,  the  body  after  impact  moves 
along  the  plane  with  the  velocity  v  sin  a. 

If  a  =  0,  so  that  the  impact  is  direct,  we  have  from  (4), 
vx  =  ev ;  i.  e.,  after  the  impact  the  body  rebounded  along 
its  former  course  with  e  times  its  former  velocity. 

If  cc  =  0,  and  e  =  0,  then  from  (4),  vx  =  0,  and  the 
body  is  brought  to  rest  by  the  impact. 

Sch. — Of  course  the  results  of  this  article  are  applicable 
to  cases  of  impact  on  any  smooth  surface,  by  substituting 
for  the  plane  on  which  the  impact  has  been  supposed  to 


382        OBLIQUE  IMPACT   OF  TWO   SMOOTH  SPHERES. 

take  place  the  plane  which  is  tangent  to  the  surface  at  the 
point  of  impact. 

'210.  Oblique  Irhpact  of  Two  Smooth  Spheres. — 

^  Two  smooth  spheres,  moving  in  given  directions  and  with 
<fiven  velocities,  impinge ;  to  determine  the  impulse  and  the 
subsequent  motion. 

Let  the  masses 
of  the  spheres  be 
M9  M' ,  their  cen- 
tres C,  C;  their 
velocities  before 
impact  v  and  v'9 
and    after    impact  Fig. 88 

v,  and  vt'.  Let  ED  be  the  line  which  joins  their  centres  at 
the  instant  of  impact  (called  the  line  of  impact):  CA  and 
CB  the  directions  of  motion  of  the  impinging  sphere,  M, 
before  and  after  impact ;  and  C'A'  and  CB'  those  of  the 
other  sphere;  let  «,  a'  be  the  angles,  ACD  and  A'C  D, 
which  the  original  directions  of  motion  make  with  the  line 
of  impact ;  ft  ft  the  angles,  BCD  and  B'C'D,  which  their 
directions  make  after  the  impact. 

It  is  evident  that,  since  the  spheres  are  smooth,  the 
entire  mutual  impulsive  pressure  takes  place  in  the  line 
joining  the  centres  at  the  instant  of  impact.  Let  R  be  the 
impulse,  and  e  the  coefficient  of  restitution.  Resolve  all 
the  velocities  along  the  line  of  impact  and  at  right  angles 
to  it ;  the  latter  will  not  be  affected  by  the  impact,  and  the 
former  will  be  affected  exactly  in  the  same  way  as  if  the 
impact  had  been  direct.  Hence,  since  the  velocities  in  the 
line  of  impact  are  v  cos  «,  v'  cos  a',  vx  cos  ft  v,'  cos  ft,  we 
have,  by  substituting  in  (3)  and  (4)  of  Art.  207, 

M' 
Vx  cos  (3  =  v  cos  a  —    r  (1  -fg)  (v  cos  a— v'  cos  «'),  (1) 


EXAMPLES.  383 

v/cos/3'  =  v'  cosa'  +  -r? — — ,  (1+e)  (vco$a—v'  cos  «'),  (2) 

w/j/c/j  «re  the  final  velocities  of  the  two  spheres  along  the  line 
of  impact  ED. 

Also,  from  (5)  of  Art.  206,  we  obtain  by  substitution, 

D  MM'     .  ,  ,N  ,ox 

=  if  +  jf?  ^  cos  a~v  cos  "  ^  O 

(See  Tait  and  Steele's  Dynamics  of  a  Particle,  p.  323.) 

Cor.  1. — Multiplying  (1)  by  M,  and  (2)  by  M ',  and  add- 
ing we  get 

Mvx  cos  (3  -f  M'vi  cos  j3'  =  Mv  cos  «  -f  M'v  '  cos  a',  (4) 

which   shows  that   the  momentum  of  the  system  resolved 
along  the  line  of  impact  is  the  same  after  impact  as  before. 

Cor.  2. — Subtracting  (2)  from  (1)  we  obtain, 

vx  cos  0  —  Vi  cos  (3'  =  —  e  (v  cos  a  —  v'  cos  a).     (5) 

That  is,  the  relative  velocity,  resolved  along  the  line  of 
impact,  after  impact  is  —  e  times  its  value  before. 


EXAMPLES. 

1.  A  body*  weighing  3  lbs.  moving  with  a  velocity  of 
10  ft.  per  second,  impinges  on  a  body  weighing  2  lbs.,  and 
moving  with  a  velocity  of  3  ft.  per  second  ;  find  the  com- 
mon velocity  after  impact.  Ans.  7-J-  ft.  per  second. 

2.  A  body  weighing  7  lbs.  moving  11  ft.  per  second, 
impinges  on  another  at  rest  weighing  15  lbs.;  find  the  com- 
mon velocity  after  impact.  Ans.  3 \  ft.  per  second. 

*  The  bodies  are  inelastic  unless  otherwise  stated.  The  first  27  examples  are  in 
direct  impact. 


384  EXAMPLES. 

3.  A  body  weighing  4  lbs.  moving  9  ft.  per  second, 
impinges  on  another  body  weighing  2  lbs.  and  moving  in 
the  opposite  direction  with  a  velocity  of  5  ft.  per  second ; 
find  the  common  velocity  after  impact. 

Ans.  4£  ft.  per  second. 

4.  A  body,  M' ,  weighing  5  lbs.  moving  7  ft.  per  second, 
is  impinged  upon  by  a  body,  M,  weighing  6  lbs.  and  mov- 
ing in  the  same  direction ;  after  impact  the  velocity  of  M' 
is  doubled :  find  the  velocity  of  M  before  impact. 

Ans.   19f  ft.  per  second. 

5.  Two  bodies,  weighing  2  lbs.,  and  4  lbs.,  and  moving  in 
the  same  direction  with  the  velocities  of  6  and  9  ft.  respec- 
tively, impinge  upon  each  other ;  find  their  common 
velocity  after  impact.  Ans.  8  ft.  per  second. 

6.  A  weight  of  2  lbs.,  moving  with  a  velocity  of  20  ft. 
per  second,  overtakes  one  of  5  lbs.,  moving  with  a  velocity 
of  5  ft.  per  second  ;  find  the  common  velocity  after  impact. 

Ans.  9f  ft.  per  second. 

7.  If  the  same  bodies  met  with  the  same  velocities  find 
the  common  velocity  after  impact. 

Ans.  2\  ft.  per  second  in  the  direction  of  the  first. 

8.  Two  bodies  of  different  masses,  are  moving  towards 
each  other,  with  velocities  of  10  ft.  and  12  ft.  per  second 
respectively,  and  continue  to  move  after  impact  with  a 
velocity  of  1-  2  ft.  per  second  in  the  direction  of  the  greater; 
compare  their  masses.  Ans.  As  3  to  2. 

9.  A  body  impinges  on  another  of  twice  its  mass  at  rest: 
show  that  the  impinging  body  loses  two-thirds  of  its 
velocity  by  the  impact. 

10.  Two  bodies  of  unequal  masses  moving  in  opposite 
directions  with  momenta  numerically  equal  meet ;  show 
that  the  momenta  are  numerical!  v  equal  after  impact. 


EXAMPLES.  385 

11.  A  body,  M,  weighing  10  lbs.  moving  8  ft.  per  second, 
impinges  on  J/',  weighing  6  lbs.  and  moving  in  the  same 
direction  5  ft.  per  second  ;  find  their  velocities  after  impact, 
supposing  0  =  1. 

Arts.  Velocity  of  M  =  5£ ;  velocity  of  M'  =  8}. 

12.  A  body,  M,  weighing  4  lbs.  moving  6  ft.  per  second, 
meets  M'  weighing  8  lbs.  and  moving  4  ft.  per  second; 
find  their  velocities  after  impact,  0  =  1. 

Ans.  Each  body  is  reflected  back,  M  with  a  velocity  of 
7|  andiF'  with  a  velocity  of  2f. 

13.  Two  balls,  of  4  and  6  lbs.  weight,  impinge  on  each 
other  when  moving  in  the  same  direction  with  velocities  of 
9  and  10  ft.  respectively  ;  find  their  velocities  after  impact, 
supposing  0  =  -J.  Ans.  10-08  and  9-28. 

14.  Find  the  kinetic  energy  lost  by  impact  in  example  5. 

Ans.  AV 

15.  Two  bodies  weighing  40  and  60  lbs.  and  moving  in 
the  same  direction  with  velocities  of  16  and  26  ft.  respec- 
tively, impinge  on  each  other:  find  the  loss  of  kinetic 
energy  by  impact.  Ans.  37-3. 

16.  An  arrow  shot  from  a  bow  starts  off  with  a  velocity 
of  120  ft.  per  second ;  with  what  velocity  will  an  arrow 
twice  as  heavy  leave  the  bow,  if  sent  off  with  three  times 
the  force  ?  Ans.  180  ft.  per  second. 

17.  Two  balls,  weighing  8  ozs.  and  6  ozs.  respectively, 
are  simultaneously  projected  upwards,  the  former  rises  to  a 
height  of  324  ft.  and  the  latter  to  256  ft.;  compare  the 
forces  of  projection.  Ans.  As  3  to  2. 

18.  A  freight  train,  weighing  200  tons,  and  traveling  20 
miles  per  hr.  funs  into  a  passenger  train  of  50  tons,  stand- 
ing on  the  same  track;  find  the  velocity  at  which  the 
remains  of  the  passenger  train  will  be  propelled  along  the 
track,  supposing  e  =  |.  A  ns.  19-2  miles  per  hr. 


386  EXAMPLES. 

19.  There  is  a  row  of  ten  perfectly  elastic  bodies  whose 
masses  increase  geometrically  bv  the  constant  ratio  3,  and 
the  first  impinges  on  the  second  with  the  velocity  of 
5  ft.  per  second ;  find  the  velocity  of  the  last  body. 

A  ns.  T^j  ft.  per  second. 

20.  A  body  weighing  5  lbs.  moving  with  a  velocity  of  14 
ft.  per  second,  impinges  on  a  body  weighing  3  lbs.,  and 
moving  with  a  velocity  of  8  ft.  per  second ;  find  the  veloci- 
ties after  impact  supposing  e  =  ^.  Ans.  11  and  13. 

21.  Two  bodies  are  moving  in  the  same  direction  with 
the  velocities  7  and  5  ;  and  after  impact  their  velocities 
are  5  and  6;  find  e,  and  the  ratio  of  their  masses. 

Ans.  e  =  i;  M'  =  2M. 

22.  A  body  weighing  two  lbs.  impinges  on  a  body  weighing 
one  lb. ;  e  is  J,  show  that  vx  —  %(v  +  v'),  and  that  i\'  =  v. 

23.  Two  bodies  moving  with  numerically  equal  velocities 
in  opposite  directions,  impinge  on  each  other;  the  result  is 
that  one  of  them  turns  back  with  its  original  velocity,  and 
the  other  follows  it  with  half  that  velocity ;  show  that  one 
body  is  four  times  as  heavy  as  the  other,  and  that  e  =  J. 

24.  A  strikes  B,  which  is  at  rest,  and  after  impact  the 
velocities  are  numerically  equal ;  if  r  be  the  ratio  of  B's 

2 
mass  to  A's  mass,  show  that  e  is  — -r- ,  and  that  B's  mass 

r  —  1 

is  at  least  three  times  A's  mass. 

25.  A  body  impinges  on  an  equal  body  at  rest ;  show 
that  the  kinetic  energy  before  impact  cannot  be  greater 
than  twice  the  kinetic  energy,  after  impact. 

26.  A  series  of  perfectly  elastic  balls  are  arranged  in  the 
same  straight  line ;  one  of  them  impinges  on  the  next, 
then  this  on  the  next  and  so  on ;  show  that  if  their  masses 
form  a  geometric  progression  of  which  the  common  ratio 


EXAMPLES,  38? 

is  r,  their  velocities  after  impact  form  a  geometric  progres- 
sion of  which  the  common  ratio  is -• 

r  +  1 

27.  A  ball  falls  from  rest  at  a  height  of  20  ft.  above  a 
fixed  horizontal  plane;  find  the  height  to  which  it  will 
rebound,  e  being  £,  and  g  being  32.  Ans.  11 J  feet. 

28.  A  ball  impinges  on  an  equal  ball  at  rest,  the  elas- 
ticity being  perfect ;  if  the  original  direction  of  the  strik- 
ing ball  is  inclined  at  an  angle  of  45°  to  the  straight  line 
joining  the  centres,  determine  the  angle  between  the 
directions  of  motion  of  the  striking  ball  before  and  after 
impact.  Ans.  45°. 

29.  A  ball  falls  from  a  height  h  on  a  horizontal  plane, 
and  then  rebounds;  find  the  height  to  which  it  rises  in  its 
ascent.  Ans.  eVi. 

30.  A  ball  of  mass  M,  impinges  on  a  ball  of  mass  M \  at 
rest ;  show  that  the  tangent  of  the  angle  between  the  old 
and  new  directions  of  the  motion  of  the  impinging  body  is 

1  +  e  M'  sin  2«   

2      '  M  +  M'  (sin3  a  —  e  cos2  a) ' 

31.  A  ball  of  mass  M  impinges  on  a  ball  of  mass  M1  at 

rest ;   find  the  condition  in  order  that   the   directions  of 

motion  of  the  impinging  ball  before  and  after  impact  may 

be  at  right  angles.  '  d  0  M'e  —  M 

Ans.  tair*  a  =    ,rl    ,    -^* 
M  +  M 

32.  A  ball  impinges  on  an  equal  ball  at  rest,  the  angle 
between  the  old  and  new  directions  of  motion  of  the 
impinging  ball  is  60°  ;  find  the  velocity  after  impact,  e 
being  1.  Ans.  v  sin  30c 


,o 


33.  A  ball  impinges  on  an  equal  ball  at  rest,  e  being  1  ; 
find  the  condition  under  which  the  velocities  will  be  equal 
after  impact.  Ans.  a  =  45° 


388  EXAMPLES. 

34.  A  ball  is  projected  from  the  middle  point  of  one  side 
of  a  billiard  table,  so  as  to  strike  first  an  adjacent  side,  and 
then  the  middle  point  of  the  side  opposite  to  that  from 
which  it  started ;  find  where  the  ball  must  hit  the  adjacent 
side,  its  length  being  b. 

Ans.  At  the  distance from  the  end  nearest  the 

1  -J-  $ 

opposite  side. 


CHAPTER    V. 

WORK    AND    ENERGY. 

211.  Definition  and  Measure  of  Work. —  Work  it 
Ihc  production  of  motion  against  resistance.  A  force  is  said 
to  do  work,  if  it  moves  the  body  to  which  it  is  applied  ; 
and  the  work  done  by  it  is  measured  by  the  product  of  the 
force  into  the  space  through  which  it  moves  the  body 
(Art.  101,  Rem.). 

Thus,  the  work  done  in  lifting  a  weight  through  a  ver- 
tical distance  is  proportional  to  the  weight  lifted  and 
the  vertical  distance  through  which  it  is  lifted.  The  unit 
of  work  used  in  England  and  in  this  country  is  that  which 
is  required  to  overcome  the  weight  of  a  pound  through  the 
vertical  height  of  a  foot,  and  is  called  a  foot-pound.  For 
instance,  if  a  weight  of  10  lbs.  is  raised  to  a  height  of 
5  ft.,  or  5  lbs.  raised  to  a  height  of  10  ft.,  50  foot-pounds  of 
work  must  have  been  expended  in  overcoming  the  resist- 
ance of  gravity.  Similarly,  if  it  requires  a  force  of  50  lbs. 
to  move  a  load  on  a  horizontal  plane  over  a  distance  of 
100  ft.,  5000  foot-pounds  of  work  must  have  been  done. 
If  a  carpenter  urges  forward  a  plane  through  3  ft.  with  a 
force  of  12  lbs.,  he  does  36  foot-pounds  of  work  ;  or,  if  a 
weight  of  7  lbs.  descends  through  10  ft.,  gravity  does 
70  foot-pounds  of  work  on  it. 

Hence,  the  number  of  units  of  work,  or  foot-pounds, 
necessary  to  overcome  a  constant  resistance  of  P  pounds 
through  a  distance  of  S  feet  is  equal  to  the  product  PS. 

From  this  it  appears  that,  if  the  point  of  application 
move  always  perpendicular  to  the  direction  in  which  the 
force  acts,  such  a  force  does  no  work.  Thus,  no  work  is 
done  by  gravity  in   the   case  of  a  particle  moving  on  a 


390  WORK  DONE  BY  A   FORCE. 

horizontal  plane,  and  when  a  particle  moves  on  any  smooth 
surface  no  work  is  done  by  the  force  which  the  surface 
exerts  upon  it. 

Neither  force  nor  motion  alone  is  sufficient  to  constitute 
work;  so  that  a  man  who  merely  supports  a  loud  without 
moving  it,  does  no  work,  in  the  sense  in  which  thai;  term  is 
used  mechanically,  any  more  than  a  column  does  which 
sustains  a  heavy  weight  upon  its  summit. 

If  a  body  is  moved  in  the  direction  opposite  to  that  in 
which  its  weight  acts,  the  agent  raising  it  does  work  upon 
it,  while  the  work  done  by  the  earth's  attraction  is  nega- 
tive. When  the  work  done  by  a  force  is  negative,  i.  e., 
when  the  point  of  application  moves  in  the  direction  oppo- 
site to  that  in  which  the  force  acts,  this  is  frerpiently 
expressed  by  saying  that  work  is  done  against  the  force. 
In  the  above  case  work  is  done  ly  the  force  lifting  the 
body,  and  against  the  earth's  attraction. 

212.  General  Case  of  Work  done  by  a  Force.— 

When  either  the  magnitude  or  direction  of  a  force  varies,  or 
if  both  of  them  vary,  the  work  done  by  the  force  during  any 
finite  displacement  cannot  be  defined  as  in  Art.  211.  In 
this  case  the  work  done  during  any  indefinitely  small  dis- 
placement may  be  found  by  supposing  the  magnitude  and 
direction  of  the  force  constant  during  the  displacement,  and 
finding  the  work  done  as  in  Art.  211 ;  then  taking  the  sum 
of  all  such  elements  of  work  done  during  the  consecutive 
small  displacements,  which  together  make  up  the  finite 
displacement,  we  obtain  the  whole  work  done  by  the  force 
during  such  finite  displacement. 

Thus  let  a  force,  P,  act  at  a  point,  0,  in  the  direction  OP  (Fig.  50) 
and  let  us  suppose  the  point,  0,  to  move  into  any  other  position,  A, 
very  near  0.  If  d  be  the  angle  between  the  direction,  OP,  of  the 
force  and  the  direction,  OA,  of  the  displacement  of  the  point  of  appli- 
cation, then  the  product,  P  •  OA  cos  d,  is  called  the  work  done  by  the 
force.  If  we  drop  a  perpendicular,  AN,  on  OP,  the  work  done  by  the 
force  is  also  equal  to  the  product  P  •  ON,  where  ON  is  to  be  esti- 


MEASURE   OF   WORK.  391 

fcrted  as  positive  when  in  the  direction  of  the  icrce.  If  several  forces 
aci,  the  work  done  by  each  can  be  found  in  the  same  way  ;  and  the 
su:n  of  all  these  is  the  work  done  by  the  whole  system  of  forces. 

It  appears  from  this  that  the  work  done  by  any  force  during  an 
innnitesimal  displacement  of  the  point  of  application,  is  the  product 
of  the  resolved  part  of  the  force  in  the  direction  of  the  displacement 
into  the  displacement  ;  and  this  is  the  same  as  the  virtual  moment  of 
the  force,  which  has  been  described  in  Art.  101.  In  Statics  we  are 
concerned  only  with  the  small  hypothetical  displacement  whwh  we 
give  the  point  of  application  of  the  force  in  applying  the  principle  of 
virtual  velocities.  But  in  Kinetics  the  bodies  are  in  motion  ;  the 
force  actually  displaces  its  point  of  application  in  such  a  maimer  that 
the  displacement  has  a  projection  along  the  direction  of  the  force.  If 
ds  denote  the  projection  oi;  any  elementary  arc  of  a  curve  along  the 
direction  of  P,  the  work  done  by  P  in  this  displacement  is  Pds.  The 
sum  of  all  these  elements  of  work  done  by  P  in  its  motion  over  a 
finite  space  is  the  whole  work  found  by  taking  the  integral  of  Pds 
between  proper  limits. 

Hence  generally,  if  s  be  an  arc  of  the  path  of  a  particle,  P  the 
tangential  component  of  the  forces  which  act  on  it,  the  work  done  on 
the  particle  between  any  two  points  of  its  path  is 

/Pds,  (1) 

the  integral  being  taken  between  limits  corresponding  to  the  initial 
and  final  positions  of  the  particle. 

213.  Work  on  an  Inclined   Plane. — Let   a  be   the 

inclination  of  the  plane  to  the  horizon,  W  the  weight 
moved,  s  the  distance  along  the  plane  through  which  the 
weight  is  moved.  Resolve  W  into  two  components,  one 
along  the  plane  and  the  other  perpendicular  to  it ;  the 
former,  IF  sin  a,  is  the  component  which  resists  motion 
along  the  plane.  Hence  the  amount  of  work  required  to 
draw  the  weight  up  the  plane  =  W  sin  «  •  s  =  IF x  the 
vertical  height  of  the  plane  ;  i.  e.,  the  amount  of  work 
required  is  unchanged  by  the  substitution  of  the  oblique  path 
for  the  vertical.  Hence  the  work  in  moving  a  body  up  an 
inclined  plane,  without  friction,  is  equal  to  the  product  of 
the  weight  of  the  body  by  the  vertical  height  through  which 
it  is  raised. 


392  WORK   ON  AN  INCLINED   PLANE. 

Cor.  1. — If  the  plane  be  rough,  let  y.  =  the  coefficient 
of  friction ;  then  since  the  normal  component  of  the  weight 
is  Wcos  cc,  the  resistance  of  friction  is  \iW cos  «  (Art.  92). 
The  work  required  consists  of  two  parts,  (1)  raising  the 
weight  along  the  plane,  and  (2)  overcoming  the  resistance 
of  friction  along  the  plane,  the  former  =  W  sin  a  •  s,  and 
the  latter  is  \i  W  cos  a  •  s.  Hence  the  ivhole  work  necessary 
to  move  the  weight  up  the  plane  is 

JF(sin  a  -f  y  cos  a)  s.  (1) 

Since  s  sin  a  represents  the  vertical  height  through 
which  the  weight  is  raised,  and  s  cos  a  the  horizontal  space 
through  which  it  is  drawn,  this  result  may  be  stated  thus  : 
The  work  expended  is  the  same  as  that  which  would  be 
required  to  raise  the  weight  through  the  vertical  height  of 
the  plane,  together  with  that  which  would  he  required  to 
draw  the  tody  along  the  base  of  the  plane  horizontally 
against  friction. 

Cor.  2. — If  a  body  be  dragged  through  a  space,  s,  down 
an  inclined  plane,  which  is  too  rough  for  the  body  to  slide 
down  by  itself  the  work  done  is 

W  (y  cos  a  —  sin  «)  s.  (2) 

Cor.  3. — If  h  =  the  height  of  the  inclined  plane,  and 
b  =  its  horizontal  base,  then  the  work  done  against  gravity 
to  move  the  body  up  the  plane  =  Wh  ;  and  the  work  done 
against  friction  to  move  the  body  along  the  plane,  suppos- 
ing it  to  be  horizontal,  =  lib  W.  Hence  (Cor.  1)  the  total 
work  done  is 

Wh+ybW.  (3) 

If  the  body  be  drawn  down  the  plane,  the  total  work 
expended  (Cor.  2)  is 

—  Wh  +  ubW.  (4) 


EXAMPLES.  393 

If  in  (4)  the  former  term  is  greater  than  the  latter, 
gravity  does  more  work  than  what  is  expended  on  friction, 
and  the  body  slides  down  the  plane  with  accelerated 
velocity. 

Sch.  1. — If  the  inclination  of  the  plane  is  small,  as  it  is 
in  most  cases  which  occur  in  practice,  as  in  common  roads 
and  railroads,  cos  a  may  without  any  important  error  be 
taken  as  equal  to  unity,  and  the  expression  for  the  work 
becomes  (Cors.  1  and  2) 

W  fas  ±  s  sin  a),  (5) 

the  upper  or  lower  sign  being  taken  according  as  the  body 
is  dragged  up  or  down  the  plane. 

Sch.  2. — If  the  inclination  of  the  plane  is  small,  as  in 
the  case  of  railway  gradients,  the  pressure  upon  the  plane 
will  be  very  nearly  equal  to  the  weight  of  the  body ;  and 
the  total  work  in  moving  a  body  along  an  inclined  plane 
will  be  from  (3)  and  (4), 

\aw±  wh9  (6) 

where  \ilW  is  the  work  due  to  friction  along  the  plane 
of  length  I,  and  Wh  is  the  work  due  to  gravity,  the  proper 
sign  being  taken  as  in  (5). 

EXAMPLES. 

1.  How  much  work  is  done  in  lifting  150  and  200  lbs. 
through  the  heights  of  80  and  120  ft.  respectively. 

The  work  done  =  150  x  80  +  200  x  120 
=  36000  foot-pounds,  Ans. 

2.  A  body  weighing  500  lbs.  slides  on  a  rough  horizontal 
plane,  the  coefficient  of  friction  being  0.1 ;  how  much  work 
must  be   done   against  friction  to  move  the   body  over 

ioo  a? 


394  EXAMPLES. 

Here  the  friction  is  a  force  of  50  lbs.  acting  directly 
opposite  to  the  motion  ;  hence  the  work  done  against  fric- 
tion to  move  the  body  over  100  ft.  is 

50  x  100  =  5000  foot-pounds,  Arts. 

3.  A  train  weighs  100  tons;  the  total  resistance  is  8  lbs. 
per  ton;  how  much  work  must  be  expended  in  raising  it 
to  the  top  of  an  inclined  plane  a  mile  long,  the  inclination 
of  the  plane  being  1  vertical  to  70  horizontal. 

Here  the  work  done  against  friction  (Sch.  2) 

=  800  x  5280  =  4224000  foot-pounds, 
and  the  work  done  against  gravity 

=  224000*  x  5280  x  -&  =  16896000  foot-pounds, 
so  that  the  whole  work  =  21120000  foot-pounds. 

4.  A  train  weighing  100  tons  moves  30  miles  an  hour 
along  a  horizontal  road;  the  resistances  are  8  lbs.  per  ton; 
find  the  quantity  of  work  expended  each  hour. 

Ans.  126,720000  foot-pounds. 

5.  If  25  cubic  feet  of  water  are  pumped  every  5  minutes 
from  a  mine  140  fathoms  deep,  required  the  amount  of 
work  expended  per  minute,  a  cubic  foot  of  water  weighing 
62£lbs.  Ans.  262500  foot-pounds. 

6.  How  much  work  is  done  when  an  engine  weighing 
10  tons  moves  half  a  mile  on  a  horizontal  road,  if  the 
total  resistance  is  8  lbs.  per  ton. 

Ans.  211200  foot-pounds. 

7.  If  a  weight  of  1120  lbs.  be  lifted  up  by  20  men,  20  ft. 
high,  twice  in  a  minute,  how  much  work  does  each  man 
do  per  hour  ?  Ans.  134400  foot-pounds. 

*  One  ton  being  2240  lbs.  unless  otherwise  stated. 


HORSE   POWER.  395 

8.  A  body  falls  down  the  whole  length  of  an  inclined 
plane  on  which  the  coefficient  of  friction  is  0.2.  The 
height  of  the  plane  is  10  ft.  and  the  base  30  ft.  On  reach- 
ing the  bottom  it  rolls  horizontally  on  a  plane,  having  the 
same  coefficient  of  friction.     Find  how  far  it  will  roll. 

Ans.  20  ft. 

9.  How  much  work  will  be  required  to  pump  8000  cubic 
feet  of  water  from  a  mine  whose  depth  is  500  fathoms. 

Ans.   1500000000  foot-pounds. 

10.  A  horse  draws  150  lbs.  out  of  a  well,  by  means  of  a 
rope  going  over  a  fixed  pulley,  moving  at  the  rate  of 
2J-  miles  an  hour;  how  many  units  of  work  does  this  horse 
perform  a  minute,  neglecting  friction. 

Ans.  33000  units  of  work. 

214.  Horse  Power. — It  would  be  inconvenient  to 
express  the  power  of  an  engine  in  foot-pounds,  since  this 
nnit  is  so  small ;  the  term  Horse  Power  is  therefore  used 
in  measuring  the  performance  of  steam  engines.  From 
experiments  made  by  Boulton  and  Watt  it  was  estimated 
that  a  horse  could  raise  33000  lbs.  vertically  through  one 
foot  in  one  minute.  This  estimate  is  probably  too  high  on 
the  average,  but  it  is  still  retained.  Whether  it  is  greater 
or  less  than  the  power  of  a  horse  it  matters  little,  while  it 
is  a  power  so  well  defined.  A  Horse  Poiver  therefore  means 
a  power  which  can  perform  33000  foot-pounds  of  work  in  a 
minute.  Thus,  when  we  say  that  the  actual  horse  power 
of  an  engine  is  ten,  we  mean  that  the  engine  is  able  to  per- 
form 330000  foot-pounds  of  work  per  minute. 

It  has  been  estimated  that  f  of  the  33000  foot-pounds  would  be 
about  the  work  of  a  horse  of  average  strength.  A  mule  will  perform 
|  the  work  of  a  horse.  An  ass  will  perform  about  4  the  work  of  a 
horse.  A  man  will  do  about  y1^  the  work  of  a  horse,  or  about  3300 
units  of  work  per  minute.  See  Evers'  Applied  Mech's;  also  Byrne's 
Practical  Mech's. 


396        WORK   OF  RAISING   A    SYSTEM   OF   WEIGHTS. 

215.  Work  of  Raising  a   System   of  Weights.— 

Let  P,  Q,  R,  be  any  three  weights  at  the  distances,  p,  q, 
r,  respectively  above  a  fixed  horizontal  plane.  Then  [Art. 
59  (3)]  or  (Art.  73,  Cor.  3),  the  distance  of  the  centre  of 
gravity  of  P,  Q,  R,  above  this  fixed  horizontal  plane  is 

Pp+  Qq  +  Rr  m 

P  +  Q  +  R   '  {  ' 

Now  suppose  that  the  weights  are  raised  vertically 
through  the  heights  a,  b,  c,  respectively.  Then  the  dis- 
tance of  the  centre  of  gravity  of  the  three  weights,  in  the 
new  position,  above  the  same  fixed  horizontal  plane  is 

P  (P  +  a)  +  Q  (q  +  b)  +  R  (r  +  c) 

P  +  Q  +  R 


(2) 


Subtracting  (1)  from  (2),  we  have 

Pa  +  Qb  +  Re 
P+~Q  +  R 


(3) 


for  the  vertical  distance  between  the  two  positions  of  the 
centre  of  gravity  of  the  three  bodies. 

Now  the  work  of  raising  vertically  a  weight  equal  to  the 
sum  of  P,  Q,  R,  through  the  space  denoted  by  (3)  is  the 
product  of  the  sum  of  the  weights  into  the  space,  which  is 

Pa  +  Qb  +  Re,  (4) 

but  (4)  is  the  work  of  raising  the  three  weights  P,  Q,  R, 
through  the  heights  a,  b,  c,  respectively.  In  the  same  way 
this  may  be  shown  for  any  number  of  weights. 

Hence  ivhen  several  loeights  are  raised  vertically  through 
(liferent  heights,  the  whole  work  done  is  the  same  as  that  of 
raising  a  weight  equal  to  the  sum  of  the  weights  vertically 
from  the  first  position  of  their  centre  of  gravity  to  the  last 
position.     (See  Todhunter's  Mech's,  p.  338.) 


EXAMPLES.  397 

EXAMPLES. 

1.  How  many  horse-power  would  it  take  to  raise  3  cwt 
of  coal  a  minute  from  a  pit  whose  depth  is  110  fathoms? 

Depth  =  110  x  6  =  660  feet. 

3  cwt.  =  112  x  3  =  336  lbs. 

Hence  the  work  to  be  done  in  a  minute 

=  660  x  336  ==  221760  foot-pounds. 

Therefore  the  horse-power 

=  221760  -^-  33000  =  6.72,  Ans. 

2.  Find  how  many  cubic  feet  of  water  an  engine  of 
40  horse-power  will  raise  in  an  hour  from  a  mine  80 
fathoms  deep,  supposing  a  cubic  foot  of  water  to  weigh 
1000  ozs.  k     ' 

Work  of  the  engine  per  hour  =  40  x  33000  x  60  foot- 
pounds. 

Work  expended  in  raising  one  cubic  foot  of  water 
through  80  fathoms  =  ifp  x  80  x  6  =  30000  foot- 
pounds. 

Hence  the  number  of  cubic  feet  raised  in  an  hour 

=  40  x  33000  x  60  -^  30000  =  2640,  Ans. 

3.  Find  the  horse-power  of  an  engine  which  is  to  move 
at  the  rate  of  20  miles  an  hour  up  an  incline  which  rises 
1  foot  in  100,  the  weight  of  the  engine  and  load  being 
60  tons,  and  the  resistance  from  friction  12  lbs.  per  ton. 

The  horizontal  space  passed  over  in  a  minute  =  1760  ft.; 
the  vertical  space  is  one-hundredth  of  this  =  17.60  ft. 
Hence  from  (6)  of  Art.  213,  we  have 

12  x  1 760  x  60  +  60  x  2240  x  1 7.6  =  1 760  x  2064  foot-pounds. 


398  EXAMPLES. 

Therefore  the  horse-power 

=  1760  x  2064  -f-  33000  =  110.08,  Am. 

4.  A  well  is  to  be  dug  20  ft.  deep,  and  4  ft.  in  diameter ; 
find  the  work  in  raising  the  material,  supposing  that  a 
cubic  foot  of  it  weighs  140  lbs. 

Here  the  weight  of  the  material  to  be  raised 

=  4t  x  20  x  140  =  140  x  80tt  lbs. 

The  work  done  is  equivalent  to  raising  this  through  the 
height  of  10  ft.  (Art.  215).     Hence  the  whole  work 

=  140  x  80tt  x  10  =  112000tt  foot-pounds,  Am. 

5.  Find  the  horse-power  of  an  engine  that  would  raise 
T  tons  of  coal  per  hour  from  a  pit  whose  depth  is  a 
fathoms. 

Work  per  minute  = — =  224a T\ 

.  • .    the  horse-power  =  ,  Ans. 

6.  Required  the  work  in  raising  water  from  three  different 
levels  whose  depths  are  a,  b,  c  fathoms  respectively  ;  from 
the  first  A,  from  the  second  B,  from  the  third  C,  cubic 
feet  of  water  are  to  be  raised  per  minute. 

Work  in  raising  water  from  the  first  level 

=  62.5.4  x  a  x  6  =  375  A-a; 

and  so  on  for  the  work  in  the  other  levels  ; 

.-.     work  per  min.  =  375  (A-a-\-B-b-\-  C-c)  foot-pounds. 

7.  Find  the  horse-power  of  an  engine  which  draws  a 
load  of  T  tons  along  a  level  road  at  the  rate  of  m  miles 


EXAMPLES.  399 

an  hour,  the  friction   being  p  pounds  per  ton,  all  other 
resistances  being  neglected. 
Work  of  the  engine  per  minute 

_,  5280  m       QQ 
=  Tp     6Q      =  88  Tpm. 


m  __  88  Tpm  __  8  Tpm 
33000    '  "    3000 


Ans. 


8.  Required  the  number  of  horse-power  to  raise  2200 
cubic  ft.  of  water  an  hour,  from  a  mine  whose  depth  is  63 
fathoms.  Arts,  26 J. 

9.  What  weight  of  coal  will  an  engine  of  4  horse-power 
raise^in  one  hour  from  a  pit  whose  depth  is  200  ft.  ? 

Ans.  39600  lbs. 

10.  In  what  time  will  an  engine  of  10  horse-power  raise 
5  tons  of  material  from  the  depth  of  132  ft.? 

Ans.  4*48  minutes. 

11.  How  many  cubic  feet  of  water  will  an  engine  of  36 
horse-power  raise  in  an  hour  from  a  mine  whose  depth  is  40 
fathoms  ?  Ans.  4752  cubic  feet. 

12.  The  piston  of  a  steam  engine  is  15  ins.  in  diameter  ; 
its  stroke  is  2£  ft.  long ;  it  makes  40  strokes  per  minute ; 
the  mean  pressure  of  the  steam  on  it  is  15  lbs.  per  square 
inch;  what  number  of  foot-pounds  is  done  by  the  steam 
per  minute,  and  what  is  the  horse-power  of  the  engine  ? 

Ans.  265072-5  foot-pounds  ;  8-03  H.-P. 

13.  A  weight  of  \\  tons  is  to  be  raised  from  a  depth  of 
50  fathoms  in  one  minute;  determine  the  horse-power  of 
the  engine  capable  of  doing  the  work. 

Ans.  30TyH.-P. 

*  The  letters  H.-P.  are  often  used  as  abbreviations  of  the  words  horse-power. 


400  MODULUS   OF  A    MACHINE. 

14.  The  resistance  to  the  motion  of  a  certain  body  is 
440  lbs.;  how  many  foot-pounds  must  be  expended  in 
making  this  body  move  over  30  miles  in  one  hour?  What 
must  be  the  horse-power  of  an  engine  that  does  the  same 
number  of  foot-pounds  in  the  same  time  ? 

Ans.  69696000  foot-pounds;  35^  H.-R 

15.  An  engine  draws  a  load  of  60  tons  at  the  rate  of  20 
miles  an  hour;  the  resistances  are  at  the  rate  of  8  lbs.  per 
ton  ;  find  the  horse-power  of  the  engine.  Ans.  25-6. 

16.  How  many  cubic  feet  of  water  will  an  engine  of  250 
horse-power  raise  per  minute  from  a  depth  of  200  fathoms  ? 

Ans.  110  cubic  ft. 

17.  There  is  a  mine  with  three  shafts  which  are  respec- 
tively 300,  450,  and  500  ft.  deep ;  it  is  required  to  raise 
from  the  first  80,  from  the  second  60,  from  the  third  40 
cubic  ft.  of  water  per  minute;  find  the  horse-power  of  the 
engine.  Ans.  134f|-. 

216.  Modulus*  of  a  Machine. — The  whole  work  per- 
formed by  a  machine  consists  of  two  parts,  the  useful  work 
and  the  lost  work.  The  useful  work  is  that  which  the 
machine  is  designed  to  produce,  or  it  is  that  which  is 
employed  in  overcoming  useful  resistances  ;  the  lost  work 
is  that  which  is  not  wanted,  but  which  is  unavoidably 
produced  or  it  is  that  which  is  spent  in  overcoming  waste* 
ful  resistances.  For  instance  in  drawing  a  train  of  cars,  the 
useful  work  is  performed  in  moving  the  train,  but  the  lost 
work  is  that  which  is  done  in  overcoming  the  friction  of 
the  train,  the  resistance  of  gravity  on  up  grades,  the  resist- 
ance of  the  air,  etc.  The  work  applied  to  a  machine  is 
equal  to  the  whole  work  done  by  the  machine,  both  useful 
and  lost,  therefore  the  useful  work  is  always  less  than  the 
work  applied  to  the  machine. 

*  Sometimes  called  Efficiency.    (Art.  108.) 


EXAMPLES.  401 

The  Modulus  of  a  machine  is  the  ratio  of  the  useful  work 
done  to  the  work  applied,  Thus,  if  the  work  applied  to  an 
engine  be  40  horse-power,  and  the  engine  delivers  only  30 
horse-power,  the  modulus  is  J,  i.  e.,  one-quarter  of  the  work 
applied  to  the  machine  is  lost  by  friction,  etc. 

Let  W  be  the  work  applied  to  the  machine,  Wu  the  use- 
ful work,  and  m  the  modulus.  Then  we  have  from  the 
above  definition 

W 
m  =  if  (1) 

If  a  machine  were  perfect,  i.  e.,  if  there  were  no  lost  work, 
the  modulus  would  be  unity;  but  in  every  machine,  some 
of  the  work  is  lost  in  overcoming  wasteful  resistances, 
so  that  the  modulus  is  always  less  than  unity  ;  and  it  is  of 
course  the  object  of  inventors  and  improvers  to  bring  this 
fraction  as  near  to  unity  as  possible. 

EXAMPLES. 

1.  Ah  engine,  of  A7"  effective  horse-power,  is  found  to 
pump  A  cubic  ft.  of  water  per  min.,  from  a  mine  a  fathoms 
deep  ;  find  the  modulus  of  the  pumps. 

Work  of  the  engine  per  min.  =  33000  N  H.-P. 

The  useful  work,  or  work  expended  in  pumping  water, 

=  62-5  A  x  Ga  =  375  A-a. 

hence  from  (1)  we  have 

M  _  375  A- a  _    A- a       . 
M  ~  33000  AT  -  88^'  An8m 

2.  There  were  A  cubic  ft.  of  water  in  a  mine  whose  depth 
is  a  fathoms,  when  an  engine  of  N  horse-poAver  began  to 
work  the  pump:  the  water  continued  to  flow  into  the  mine 
at  the  rate  of  B. cubic  ft.  per  minute;  required  the  time 


402  EXAMPLES. 

in  which  the  mine  would  be  cleared  of  water,  the  modulus 
of  the  pump  being  m. 

Let  x  =  the  number  of  minutes  to  clear  the  mine  ot 
water.     Then 

weight  of  water  to  be  pumped  =  62-5  (A  -f-  Bx) ; 

»vork  in  pumping  water  =  375#  (A  -f  Bx)  foot-pounds; 

effective  work  of  the  engine  =  m-  ^330002;; 

.  • .    33000  mJSTx  =  375a  (A  +  Bx) ; 

A-a  . 

*'•    X~  SSmN-B-a'  An8' 

3.  An  engine  has  a  6  foot  cylinder ;  the  shaft  makes  30 
revolutions  per  minute,  the  average  steam  pressure  is  25 
lbs.  per  square  inch  ;  required  the  horse-power  when  the 
area  of  the  piston  is  1800  square  inches,  the  modulus  of 
the  engine  being  \%. 

Work  done  in  one  minute  =  1800  x  25  x  6  x-2  x  30 
foot-pounds.  We  multiply  by  twice  the  length  of  the 
stroke,  because  the  piston  is  driven  both  up  and  down  in 
one  revolution  of  the  shaft. 

The  effective  horse-power  =  J-& <Lo*||*i2*iLo  x  11 

=  450,  Ans. 

4.  The  diameter  of  the  piston  of  a  steam  engine  is  60 
ins.;  it  makes  11  strokes  per  minute;  the  length  of  each 
stroke  is  8  ft.;  the  mean  pressure  per  square  in.  is  15  lbs.; 
required  the  number  of  cubic  ft.  of  water  it  will  raise  per 
hour  from  a  depth  of  50  fathoms,  the  modulus  of  the 
engine  being  0«65. 

The  number  of  foot-pounds  of  useful  work  done  in  one  hour  and 
spent  in  raising  water  =  n  x  302  x  8  x  15  x  11  x  60  x  0  •  65,  therefore,  etc 

Ans.  7763  cubic  ft 


EXAMPLES.  403 

5.  An  engine  is  required  to  pump  1000000  gallons  of 
Mater  every  12  hours,  from  a  mine  132  fathoms  deep  ;  find 
the  horse-power  if  the  modulus  be  fi,  and  a  gallon  of 
water  weighs  10  lbs.  Am.  363T\  II.-P. 

6.  What  must  be  the  horse-power  of  an  engine  working 
e  hours  per  day,  to  supply  n  families  with  g  gallons  of 
water  each  per  day,  supposing  the  water  to  be  raised  to  the 
mean  height  of  li  feet,  and  that  a  gallon  of  water  weighs  10 
lbs.,  the  modulus  being  m.  .  ngh        „  p 

198000  em 

7.  Water  is  to  be  raised  from  a  mine  at  two  different 
levels,  viz.,  50  and  80  fathoms,  from  the  former  30  cubic  ft., 
and  from  the  latter  15  cubic  ft.  per  minute  ;  find  the  horse- 
power of  the  machinery  that  will  be  required,  assuming 
the  modulus  to  be  0-  6.  Ans.  51-14  H.-P. 

8.  The  diameter  of  the  piston  of  an  engine  is  80  ins.,  the 
mean  pressure  of  the  steam  is  12  lbs.  per  square  inch,  the 
length  of  the  stroke  is  10  ft.,  the  number  of  strokes  made 
per  minute  is  11 ;  how  many  cubic  ft.  of  water  will  it  raise 
per  minute  from  a  depth  of  250  fathoms,  its  modulus  being 
0*0?  Ans.  42-46  cubic  ft. 

9.  If  the  engine  in  the  last  example  had  raised  55  cubic 
ft.  of  water  per  minute  from  a  depth  of  250  fathoms,  what 
would  have  been  its  modulus  ?  Ans.  0-  7771. 

10.  How  many  strokes  per  minute  must  the  engine  in 
Ex.  8  make  in  order  to  raise  15  cubic  ft.  of  water  per 
minute  from  the  given  depth  ?  Ans.  4. 

11.  What  must  be  the  length  of  the  stroke  of  an  engine 
whose  modulus  is  0-  65,  and  whose  other  dimensions  and 
conditions  of  working  are  the  same  as  in  Ex.  8,  if  they  both 
do  the  same  quantity  of  useful  work  ?  Ans.  9-23  ft. 


404  KINETIC  AND   POTENTIAL    ENERGY. 

217.  Kinetic  and  Potential  Energy.  Stored 
Work. — The  energy  of  a  body  means  its  power  of  doing 
work;  and  the  total  amount  of  energy  possessed  by  the  body 
is  measured  by  the  total  amount  of  work  which  it  is  capable 
of  doing  in  passing  from  its  present  condition  to  some 
standard  condition. 

Every  moving  body  possesses  energy,  for  it  can  be  made 
to  do  work  by  parting  with  its  velocity.  The  velocity  of 
the  body  may  be  used  for  causing  it  to  ascend  vertically 
against  the  attraction  of  the  earth,  i.  e,9  to  do  work  against 
the  resistance  of  gravity.  A  cannon  ball  in  motion  can 
penetrate  a  resisting  body ;  water  flowing  against  a  water- 
wheel  will  turn  the  wheel ;  the  moving  air  drives  the  ship 
through  the  water.  Wherever  we  find  matter  in  motion 
we  have  a  certain  amount  of  energy. 

Energy,  as  known  to  us,  belongs  to  one  or  the  other  of 
two  classes,  to  which  the  names  kinetic*  energy  and 
potential  energy  are  given. 

Kinetic  energy  is  energy  that  a  body  possesses  in  virtue  of 
its  being  in  motion.  It  is  energy  actually  in  use,  energy 
that  is  constantly  being  spent.  The  energy  of  a  bullet  in 
motion,  or  of  a  fly-wheel  revolving  rapidly,  or  of  a  pile- 
driver  just  before  it  strikes  the  pile,  are  examples  of  kinetic 
energy.  The  work  done  by  a  force  on  a  body  free  to  move, 
exerted  through  a  given  distance,  is  always  equal  to  the 
corresponding  increase  of  kinetic  energy  [Art.  189  (3)].  If 
a  mass,  m,  is  moving  with  a  velocity,  v,  its  kinetic  energy 
is  ^mv2  [(3)  of  Art.  189].  If  this  velocity  be  generated  by  a 
constant  force,  P,  acting  through  a  space,  s,  we  have, 
(Art.  211) 

Ps  =  hnv2,  (!) 

that  is,  the  work  done  on  the  body  is  the  exact  equivalent 
of   the   kinetic   energy,    and   the   kinetic  energy  is  recon- 

*  Called  also  actual  energy,  or  enerau  of  motion. 


KINETIC  AND   POTENTIAL   ENERGY.  4U5 

vertible  into  the  work ;  and  the  exact  amount  of  work 
which  the  mass  m,  with  a  velocity  v,  can  do  against  resist- 
ance before  its  motion  is  completely  destroyed  is  l/nv2. 
This  is  called  stored  work,*  and  is  the  amount  of  work  that 
any  opposing  force,  P,  will  have  to  do  on  the  body  before 
bringing  it  to  rest.  Thus,  when  a  heavy  fly-wheel  is  in 
rapid  motion,  a  considerable  portion  of  the  work  of  the 
engine  must  have  gone  to  produce  this  motion  ;  and  before 
the  engine  can  come  to  a  state  of  rest  all  the  work  stored 
up  in  the  fly-wheel,  as  well  as  in  the  other  parts  of  the 
machine,  must  be  destroyed.  In  this  way  a  fly-wheel  acts 
as  a  reservoir  of  work. 

If  a  body  of  mass  m,  moving  through  a  space  s,  change 
its  velocity  from  v  to  v0  the  work  done  on  the  body  as  it 
moves  through  that  space,  (Art.  189),  is 

|a»  (*-«•$  (2) 

If  the  body  is  not  perfectly  free,  i.  e.9  if  there  is  one  force 
urging  the  body  on,  and  another  force  resisting  the  body, 
the  kinetic  energy,  \mv2,  gives  the  excess  of  the  work  done 
by  the  former  force  over  that  done  by  the  latter  force. 
Thus,  when  the  resistance  of  friction  is  overcome,  the 
moving  forces  do  work  in  overcoming  this  resistance,  and 
all  the  work  done,  in  excess  of  that,  is  stored  in  the  moving 
mass. 

Potential  energy  is  energy  that  a  body  possesses  in  virtue 
of  its  position.  The  energy  of  a  bent  watch-spring,  which 
does  work  in  uncoiling ;  the  energy  of  a  weight  raised 
above  the  earth,  as  the  weight  of  a  clock  which  does  work 
in  falling  ;  the  energy  of  compressed  air,  as  in  the  air-gun, 
or  in  an  air-brake  on  a  locomotive,  which  does  work  in 
expanding;  the  energy  of  water  stored  in  a  mill-dam,  and 
of  steam  in  a  boiler,  are  all  examples  of  potential  energy. 

*  Called  also  accumulated  work,  See  Todhunter's  Mechs.,  also  stored  energy  and 
not  work.    Browne's  Mechanics,  p.  178. 


40  G  EXAMPLES. 

Such  energy  may  or  may  not  be  called  into  action,  it  maj 
be  dormant  for  years;  the  power  exists,  but  the  action  will 
begin  only  when  the  weight,  or  the  water,  or  the  steam  is 
released.  Hence  the  word  potential,  is  significant,  as 
expressing  that  the  energy  is  in  existence,  and  that  a  new 
power  has  been  conferred  upon  it  by  the  act  of  raising  or 
confining  it. 

For  example  suppose  a  weight  of  1  lb.  be  projected 
vertically  upwards  with  a  velocity  of  32-2  ft.  per  second. 
The  energy  imparted  to  the  body  will  carry  it  to  a  height 
of  16-1  ft,  when  it  will  cease  to  have  any  velocity.  The 
whole  of  its  kinetic  energy  will  have  been  expended;  but 
the  body  will  have  acquired  potential  energy  instead  ;  i.  e., 
the  kinetic  energy  of  the  body  will  all  have  been  converted 
into  potential  energy,  which,  if  the  weight  be  lodged  for 
any  time,  is  stored  up  and  ready  to  be  freed  whenever  the 
body  shall  be  permitted  to  fall,  and  bring  it  back  to  its 
starting  point  with  the  velocity  of  32-  2  ft.  per  second ;  and 
thus  the  body  will  reacquire  the  kinet;c  energy  which  it 
originally  received.  Hence  kinetic  energy  and  potential 
energy  are  mutually  convertible. 

Let  h  be  the  height  through  which  a  body  must  fall  to 
acquire  the  velocity  v,  m  and  W  the  mass  and  weight, 
respectively.  Then  since  v2  =  2gh,  we  have,  for  the  stored 
work, 

imv>  =  ^  =  ^.2gh=  Wh.  (3) 

Hence  we  may  say  that  the  work  stored  in  a  moving  body 
is  measured  by  the  product  of  the  iveight  of  the  body  into  the 
height  through  tvhich  it  must  fall  to  acquire  the  velocity. 

EXAMPLES. 

1.  Let  a  bullet  leave  the  barrel  of  a  gun  with  the  velocity 
of  1000  ft.  per  second,  and  suppose  it  to  weigh  2  ozs. ;  find 


EXAMPLES.  40? 

the  work  stored  up  in  the  bullet,  and  the  height  from  which 
it  must  fall  to  acquire  that  velocity. 
Here  we  have  from  (3)  for  the  stored  work 

(1000)2  =  Wh 


2  x  lQg 

=  1941  foot-pounds. 
.-.    h  =  15528  feet. 

2.  A  ball  weighing  w  lbs.  is  projected  along  a  horizontal 
plane  with  the  velocity  of  v  ft.  per  second  ;  what  space,  s, 
will  the  ball  move  over  before  it  comes  to  a  state  of  rest, 
the  coefficient  of  friction  being/? 

Here  the  resistance  of  friction  is  fw,  which  acts  directly 

opposite  to  the  motion,  therefore  the  work  done  by  friction 

while  the  body  moves  over  s  feet  =  fws ;  the  work  stored 

wtfi 
up  in  the  ball  =  ^mv%  =  —  ;  therefore  from  (1)  we  have 

/wv2  v* 

3.  A  railway  train,  weighing  Ttons,  has  a  velocity  of  v 
ft.  per  second  when  the  steam  is  turned  off  ;  what  distance, 
s,  will  the  train  have  moved  on  a  level  rail,  whose  friction 
is  jo  lbs.  per  ton,  when  the  velocity  is  v0  ft.  per  second  ? 

Here  the  work  done  by  friction  =  pTs\  hence  from  (2) 
we  have 

m        t    22±0T,»         2, 

p    =  * " 7" Y~ (   " Vo) ; 

.      „  _  1120  (t*  -  V) 

•    •       o    —  ■  "• 

gp 

4.  A  train  of  T  tons  descends  an  incline  of  5  ft.  in 
length,  having  a  total  rise  of  //  ft.;  what  will  be  the  velocity, 
V,  acquired  by  the  train,  the  friction  beings  lbs.  per  ton? 


408  KINETIC  ENERGY   OF  A    RIGID   BODY. 

Here  we  have  (Art.  213,  Sell.  2),  the  work  done  on  the 
train  =  the  work  of  gravity  —  the  work  of  friction 

=  2240  Th  —pTs; 

which  is  equal  to  the  work  stored  up  in  the  train.     Hence 

2240  Tv2 


*9 


=  2240  Th  —pTs; 


.-.    v  =  Vtyh  —  ttVo^s- 

5.  If  the  velocity  of  the  train  in  the  last  example  be 
v0  ft.  per  second  when  the  steam  is  turned  off,  what  will  be 
its  velocity,  v,  when  it  reaches  the  bottom  of  the  incline? 

Ans.  v  =  Vv0*  +  tyh  —  TTTsgps- 

6.  A  body  weighing  40  lbs.  is  projected  along  a  rough 
horizontal  plane  with  a  velocity  of  150  ft.  per  sec;  the 
coefficient  of  friction  is  J;  find  the  work  done  against 
friction  in  five  seconds.  Ans.  3500  foot-pounds. 

7.  Find  the  work  accumulated  in  a  body  which  weighs 
300  lbs.  and  has  a  velocity  of  64  ft.  per  second. 

Ans.  19200  foot-pounds. 

218.  Kinetic  Energy  of  a  Rigid  Body  revolving 
round  an  Axis. — Let  m  be  the  mass  of  any  particle  of 
the  body  at  the  distance  r  from  the  axis,  and  let  w  be  the 
angular  velocity,  which  will  be  the  same  for  every  particle, 
since  the  body  is  rigid;  then  the  kinetic  energy  of  m  = 
tyn  (rw)2.  The  kinetic  energy  of  the  whole  body  will  be 
found  by  taking  the  sum  of  these  expressions  for  every 
particle  of  the  body.    Hence  it  may  be  written 

2  $mrW  =  ~  2  mr2.  (1) 

4 


EXAMPLES,  409 

£  mr2  is  called  the  moment  of  inertia  of  the  body  about  the 
axis,  and  will  be  explained  in  the  next  chapter. 

Hence  the  kinetic  energy  of  any  rotating  body  =  £lw2, 
where  I  is  the  moment  of  inertia  round  the  axis,  and  w  the 
angular  velocity. 

In   the  case  of  a  fly-wheel,  it  is  sufficient  in  practice  to 

treat  the  whole  weight  as  distributed  uniformly  along  the 

circumference  of  the  circle  described  by  the  mean  radius 

of  the  rim.     Let  r  be  this  radius ;    then  the  moment  of 

inertia  of  any  particle  of  the  wheel  =  mr2,  and  the  moment 

of  inertia  of  the  whole  wheel  =  Mr2,  where  M  is  the  total 

w2 
mass.     Hence,  substituting  in  (1)  we  have  —  Mr2,  which 

is  the  kinetic  energy  of  the  fly-wheel. 

EXAMPLES. 

1.  Two  equal  particles  are  made  to  revolve  on  a  vertical 
axis  at  the  distances  of  a  and  b  feet  from  it ;  required  the 
point  wrhere  the  two  particles  must  be  collected  so  that  the 
work  may  not  be  altered. 

Let  m  =  the  mass  of  each  particle,  h  =  the  distance  of 
the  required  point  from  the  axis,  and  w  =  the  angular  veloc- 
ity ;  then  we  have 

Work  stored  in  both  particles  =  \m  («w)2  +  \m  (bu)2 ; 

Work  stored  in  both  particles  collected  at  point  =  m  (k(*>)2; 

.*.    in  (ku)2  =  \m  (acj)2  +  \m  (#w)2; 


.-.     h  =  Vi{a2  +  b2). 

This  point  is  called   the   centre  of  gyration.     (See  next 
chapter.) 

2.  The  weight  of  a  fly-wheel  is  w  lbs.,  the  wheel  makes 
n  revolutions  per  minute,  the  diameter  is  2r  feet,  diameter 


410  EXAMPLES. 

of  axle  a  inches,  and  the  coefficient  of  friction  on  the  axle 
f;  how  many  revolutions,  x,  will  the  wheel  make  before  it 
stops  ? 


Work  stored  in  the  wheel  =  —  \--p~ )  t*2, 

%q  \  60  / 


Work  done  by  friction  in  x  revolutions 
and  when  the  wheel  stops,  we  have 

loOfag 

3.  Required  the  number  of  strokes,  x,  which  the  fly-wheel 
in  the  last  example,  will  give  to  a  forge  hammer  whose 
weight  is  W  lbs.  and  lift  h  feet,  supposing  the  hammer  to 
make  one  lift  for  every  revolution  of  the  wheel. 

Here  the  work  due  to  raising  hammer  =  Whx.        .  • .  &c. 

w  -n'hih'2 


'  ~~  150#  (12  Wh  +  irafw) 

4.  The  weight  of  a  fly-wheel  is  8000  lbs.,  the  diameter 
20  feet,  diameter  of  axle  14  inches,  coefficient  of  friction 
0.2  ;  if  the  wheel  is  separated  from  the  engine  when  mak- 
ing 27  revolutions  per  minute,  find  how  many  revolutions 
it  will  make  before  it  stops  (g  taken  ==  32.2). 

Ans.  16.9  revolutions. 


EXAMPLES.  411 

219.  Force  of  a  Blow. — In  order  to  express  the 
amount  of  force  between  the  face  of  a  hammer,  for  in' 
stance,  and  the  head  of  a  nail,  we  must  consider  what 
weiglit  must  be  laid  upon  the  head  of  the  nail  to  force  it 
into  the  wood.  A  nail  requires  a  large  force  to  pull  it  out, 
when  friction  alone  is  retaining  it,  and  to  force  it  in  must 
of  course  require  a  still  larger  force. 

Now  the  head  of  the  hammer,  when  it  delivers  a  blow 
upon  the  head  of  the  nail,  must  be  capable  of  developing  a 
force  equal  for  a  short  time  to  the  continued  pressure  that 
would  be  produced  by  a  very  heavy  load.  Hence,  the  effect 
of  the  hammer  may  be  explained  by  the  principles  of  energy. 
When  the  hammer  is  in  motion  it  has  a  quantity  of  kinetic 
energy  stored  up  in  it,  and  when  it  comes  in  contact  with 
the  nail  this  energy  is  instantly  converted  into  work  which 
forces  the  nail  into  the  wood. 

EXAMP  LES. 

1.  Suppose  that  a  hammer  weighs  1  lb.,  and  that  it  is 
impelled  downwards  by  the  arm  with  considerable  force,  so 
that,  at  the  instant  the  head  of  the  hammer  reaches  the 
nail,  it  is  moving  with  a  velocity  of  20  ft.  per  second  ;  find 
the  force  which  the  hammer  exerts  on  the  nail  if  it  is 
driven  into  the  wood  one-tenth  of  an  inch. 

Let  P  be  the  force  which  the  hammer  exerts  on  the  nail, 
then  the  work  done  in  forcing  the  nail  into  the  wood  = 
P  x  xlif,  and  the  energy  stored  up  in  the  hammer 

=  \mv*  =  ^  =  6.2. 

64 

Since  the  work  done  in  forcing  the  nail  into  the  wood 
must  be  equal  to  all  the  work  stored  in  the  hammer,  (Art. 
217),  we  have 

^  =  6.2;    .-.    P=  744  lbs. 


412  EXAMPLES. 

Hence  the  force  which  the  hammer  exerts  on  the  head  of 
the  nail  is  at  least  744  lbs. 

2.  If  the  hammer  in  the  last  example  forces  the  nail  into 
the  wood  only  0.01  of  an  inch,  find  the  force  exerted  on 
the  nail.  Ans.  7440  lbs. 

Hence,  we  see  that,  according  as  the  wood  is  harder,  i.  e.,  accord 
ing  as  the  nail  enters  less  at  each  stroke,  the  force  of  the  blow 
becomes  greater.  So  that  when  we  speak  of  the  "  force  of  a  blow,' 
we  must  specify  how  soon  the  body  giving  the  blow  will  come  to 
rest,  otherwise  the  term  is  meaningless.  Thus,  suppose  a  ball  of 
100  lbs.  weight  have  a  velocity  that  will  cause  it  to  ascend  1000  ft. ; 
if  the  ball  is  to  be  stopped  at  the  end  of  1000  ft.,  a  force  of  100  lbs. 
will  do  it ;  but  if  it  is  to  be  stopped  at  the  end  of  one  foot,  it  will 
need  a  force  of  100000  lbs.  to  do  it ;  and  to  stop  it  at  the  end  of  one 
inch  will  require  a  force  of  1200000  lbs.,  and  so  on. 

220.  Work  of  a  Water-Fall. — When  water  or  any 
body  falls  from  a  given  height,  it  is  plain  that  the  work 
which  is  stored  up  in  it,  and  which  it  is  capable  of  doing,  is 
equal  to  that  which  would  be  required  to  raise  it  to  the 
height  from  which  it  has  fallen ;  i.  e.,  if  1  lb.  of  water 
descend  through  1  foot  it  must  accumulate  as  much  work 
as  would  be  required  to  raise  it  through  1  foot.  Hence 
when  a  fall  of  water  is  employed  to  drive  a  water-wheel,  or 
any  other  hydraulic  machine,  whose  modulus  is  given,  the 
work  done  upon  the  machine  is  equal  to  the  weight  of  the 
water  in  pounds  x  its  fall  in  feet  x  the  modulus  of  the 
machine. 

examples. 

1.  The  breadth  of  a  stream  is  b  feet,  depth  a  feet,  mean 
velocity  v  feet  per  minute,  and  the  height  of  the  fall  k  feet ; 
find  (1)  the  horse-power,  N,  of  the  water-wheel  whose 
modulus  is  m,  and  (2)  find  the  number  of  cubic  feet,  A, 
which  the  wheel  will  pump  per  minute  from  the  bottom  of 
che  fall  to  the  height  of  hx  feet. 


EXAMPLES.  413 

Weight  of  water  going  over  the  fall  per  min.  =  62.5  abv. 

.*.    Work  of  wheel  per  min.  =  (12.5  abvhm.  (1) 

_T       62.5  abchm 
•*•    N=  -33000-  (2> 

Work  in  pumping  water  per  min.  =  62.5  A7iy; 

which   must  =  the  work  of   the  wheel   per  min.;    hence 
from  (1)  we  have 

62.5  Ahx  =  62.5  dbvhm\ 

abvhm 


»i 


(3) 


2.  The  mean  section  of  a  stream  is  5  ft.  by  2  ft. ;  its 
mean  velocity  is  35  ft.  per  minute  ;  there  is  a  fall  of  13  ft. 
on  this  stream,  at  which  is  erected  a  water-wheel  whose 
modulus  is  0.65  ;  find  the  horse-power  of  the  wheel. 

Ans.  5.6  H.-P. 

3.  In  how  many  hours  would  the  wheel  in  Ex.  2  grind 
8000  bushels  of  wheat,  supposing  each  horse-power  to  grind 
1  bushel  per  hour?  Ans.  1428-f  hours. 

4.  How  many  cubic  feet  of  water  must  be  made  to 
descend  the  fall  per  minute  in  Ex.  2,  that  the  wheel  may 
grind  at  the  rate  of  28  bushels  per  hour  ? 

Ans.  1*750  cu.  ft. 

5.  Given  the  stream  in  Ex.  2,  what  must  be  the  height 
of  the  fall  to  grind  10  bushels  per  hour,  if  the  modulus  of 
the  wheel  is  0.4  ?  Ans.  37.7  feet. 

6.  Find  the  useful  horse-power  of  a  water-wheel,  sup- 
posing the  stream  to  be  5  ft.  broad  and  2  ft.  deep,  and  to 
flow  with  a  velocity  of  30  ft.  per  minute;  the  height  of  the 
fall  being  14  ft.,  and  the  modulus  of  the  machine  0.65. 

Ans.  5.2  nearly. 


414  EXAMPLES. 

221.  The  Duty  of  an  Engine. —  Tlie  duty  of  an  engirt* 
is  the  number  of  units  of  work  which  it  is  capable  of  doing 
by  burning  a  given  quantity  of  fuel. — It  has  been  found  by 
experiment  that,  whatever  may  be  the  pressure  at  which 
the  steam  is  formed,  the  quantity  of  fuel  necessary  to 
evaporate  a  given  volume  of  water  is  always  nearly  the 
same;  hence  it  is  most  advantageous  to  employ  steam  of  a 
high  pressure.* 

In  good  ordinary  engines  the  duty  varies  between  200000  and 
500000  units  of  work  for  a  lb.  of  coal.  The  extent  to  which  the 
economy  of  fuel  may  be  carried  is  well  illustrated  by  the  engines  em- 
ployed to  drain  the  mines  in  Cornwall,  England.  In  1815,  the 
average  duty  of  these  engines  was  20  million  units  of  work  for  a 
bushel  f  of  coal ;  in  1843,  by  reason  of  successive  improvements,  the 
average  duty  had  become  60  millions,  effecting  a  saving  of  £85000 
per  annum.  It  is  stated  that  in  the  case  of  one  engine,  the  duty  was 
raised  to  125  millions.  The  duty  of  the  engine  depends  largely  on 
the  construction  of  the  boiler ;  1  lb.  of  coal  in  the  Cornish  boiler 
evaporates  11|  lbs.  of  water,  while  in  a  differently-shaped  boiler  8.7 
is  the  maximum.^ 

EXAM  PLES. 

1.  An  engine  burns  2  lbs.  of  coal  for  each  horse-power 
per  hour  ;  find  the  duty  of  the  engine  for  a  lb.  of  coal. 

Here  the  work  done  in  one  hour 

=  60  x  33000  foot-pounds ; 

therefore  the  duty  of  the  engine  =  30  x  33000  foot-pounds, 

=  990000  foot-pounds. 

2.  How  many  bushels  of  coal  must  be  expended  in  a 
day  of  24  hours  in  raising  150  cubic  ft.  of  water  per  minute 

*  See  Tate  in  Mechanics'  Magazine,  in  the  year  1841. 

t  One  bushel  of  coal  =  84  or  94  lbs.,  depending  upon  where  it  is.  Goodeve, 
p.  120. 

X  Bourne  on  the  Steam  Engine,  p.  171,  and  Fairbairn,  Useful  Information, 
D.  177. 


WORK   OF  A     VARIABLE  FORCE.  415 

from  a  depth  of  100  fathoms  ;  the  duty  of  the  engine 
being  60  millions  for  a  bushel  of  coal  ? 

Ans.  135  bushels. 

3.  A  steam  engine  is  required  to  raise  70  cubic  ft.  of 
water  per  minute  from  a  depth  of  800  ft. ;  find  how  many 
tons  of  coal  will  be  required  per  day  of  24  hours,  supposing 
the  duty  of  the  engine  to  be  250000  for  a  lb.  of  coal. 

Ans.  9  tons. 

222.  Work  of  a  Variable  Force.— When  the  force 
which  performs  work  through  a  given  space  varies,  the 
work  done  may  be  determined  by  multiplying  the  given 
space  by  the  mean  of  all  the  variable  forces. 

Let  AG  represent  the  space  in  units 
of  feet  through  which  a  variable 
force  is  exerted.  Divide  AG  into 
six  equal  parts,  and  suppose  P19  P2, 
P3,  etc.,  to  be  the  forces  in  pounds  _.  89 

applied  at  the  points  A,  B,  0,  etc., 

respectively.  At  these  points  draw  the  ordinates  yt,  y2,  yZi 
etc.,  to  represent  the  forces  which  act  at  the  points  A,  B, 
C,  etc.  Then  the  work  done  from  A  to  B  will  be  equal  to 
the  space,  AB,  multiplied  by  the  mean  of  the  forces  P0 
and  P19  i.  e.,  the  work  will  be  represented  by  the  area  of 
the  surface  AabB.  In  like  manner  the  work  done  from 
B  to  C  will  be  represented  by  the  area  BbcC,  and  so 
on  ;  so  that  the  work  done  through  the  whole  space,  AG, 
by  a  force  which  varies  continuously,  will  be  represented  by 
the  area  AagG.  This  area  may  be  found  approximately  by 
the  ordinary  rule  of  Mensuration  for  the  area  of  a  curved 
surface  with  equidistant  ordinates,  or  more  accurately  by 
Simpson's*  rule,  the  proof  of  which  we  shall  now  give. 

223.  Simpson's   Rule.— Let  y19  y2,  y3,  etc.,  be   the 

*  Although  it  was  not  invented  by  Simpson.    See  TodhunteK 


416 


SIMPSON'S  RULE. 


equidistant  ordinates  (Fig.  89)  and  I  the  distance  between 
any  two  consecutive  ordinates;  then  by  taking  the  sum  of 
the  trapezoids,  A#oB,  BocC,  etc.,  we  have  for  the  area  of 
AagQy 

¥  {Vi  +  y%)  +  V  (1/2  +  2/3)  +  ¥  (2/3  +  yj  +  etc- 

=  ¥  (l/i  +  ^2  +  %3  +  %yA  +  2y 5  +  2y6  +  y,);  (1) 

which  is  the  ordinary  formula  of  mensuration. 

Now  it  is  evident  that  when  the  curve  is  always  concave 
to  the  line  AG  (1)  will  give  too  small  a  result,  and  if  con- 
vex it  will  give  too  large  a  result. 

Let  Fig.  90  represent  the  space  between  any  two  odd 
consecutive  ordinates,  as  Cc  and  Ee  (Fig.  89) ;  divide  CE 
into  three  equal  ps*rts,  CK  =  KL  =  LE, 
and  erect  the  ordinates  Kk  and  Id,  dividing 
the  two  trapezoids  CcdD  and  Dt/eE  into  the 
three  trapezoids  CckK,  ~KklL,  and  L?eE. 
The  sum  of  the  areas  of  these  three  trapezoids  Fig. 90 

=  iCK  (Cc  +  2Kh  +  2LZ  +  Ee) 

=  \l  (Cc  +  2K&  +  2U  +  Be),  (since  JCK  =  JCD  ==  JZ) 

=  \l  (Cc  +  4Do  +  Eg),  (since  2K&  +  2L7  =  4Do),       (2) 

which  is  a  closer  approximation  for  the  area  of  Cc^E 
than  (1). 

Now  when  the  curve  is  concave  towards  CE,  (2)  is 
smaller  than  the  area  between  CE  and  the  curve  chile;  if 
we  substitute  for  Do,  the  ordinate  Dd,  which  is  a  little 
greater  than  Do  and  which  is  given,  (2)  becomes 


\l  (Cc  +  *Dd  +  Ee), 
which  is  a  still  closer  approximation  than  (2). 


(3) 


EXAMPLES,  417 

Similarly  wo  have  for  the  areas  of  AacC  and  Ee^G, 

il  (A«  +  4IM  +  Cc),  and  Jl  (Ee  +  4F/  +  G#).      (4) 

Adding  (3)  and  (4)  together,  we  have  for  an  approximate 
value  of  the  whole  area, 

V  bfi  +  Vi  +  2  0/3  +  y8)  +  4  (y8  +  y4  +  y6)],    (5) 

which  is  Simpson's  Formula.  Hence  Simpson's  rule  for 
finding  the  area  approximately  is  the  following:  Divide  the 
abscissa,  AG,  into  an  even  number  of  equal  parts,  and  erect 
ordinates  at  the  points  of  division ;  then  add  together  the 
■first  and  last  ordinates,  twice  the  sum  of  all  the  other  odd 
ordinates,  and  four  times  the  sum  of  all  the  even  ordinates  ; 
multiply  the  sum  by  one-third  of  the  common  distance 
betioeen  any  two  adjacent  ordinates.  (See  Todhunters 
Mensuration,  also  Tate's  Geometry  and  Mensuration,  also 
Morin's  Mech's,  by  Bennett. ) 

EXAMPLES. 

1.  A  variable  force  has  acted  through  3  ft.;  the  value  of 
the  force  taken  at  seven  successive  equidistant  points, 
including  the  first  and  the  last,  is  in  lbs.  189,  151.2,  126, 
108,  94.5,  84,  75.  G  ;  find  the  whole  work  done. 

Ans.  34G.4  foot-pounds. 

2.  A  variable  force  has  acted  through  6  ft. ;  the  value  of 
the  force  taken  at  seven  successive  equidistant  points, 
including  the  first  and  the  last,  is  in  lbs.  3,  8,  15,  24,  35, 
48,  63  ;  find  the  whole  wrork  done. 

Ans.  162  foot-pounds. 

3.  A  variable  force  has  acted  through  9  ft.;  the  value  of 
the   force   taken   at   seven   successive   equidistant    points, 
including   the   first   and   the  last,  is  in  lbs.   6.082,  6.164 
6.245.  6.32a  6.403,  6.481,  6.557;   find  the  whole  work  done. 

Ans.  56.907  foot-pounds. 


418  EXAMPLES. 

Should  any  of  the  ordinates  become  zero,  it  will  not  pre- 
vent the  use  of  Simpson's  rule. 

Simpson's  rule  is  applicable  to  other  investigations  as 
well  as  to  that  of  work  done  by  a  variable  force.  For 
example,  if  we  want  the  velocity  generated  in  a  given  time 
m  a  particle  by  a  variable  force,  let  the  straight  line  AG 
represent  the  wiiole  time  during  which  the  force  acts,  and 
let  the  straight  lines  at  right  angles  to  AG  represent  the 
force  at  the  corresponding  instants ;  then  the  area  will 
represent  the  whole  space  described  in  the  given  time. 

EXAMPLES. 

1.  The  ram  of  a  pile-driving  engine  weighs  half  a  ton,* 
and  has  a  fall  of  17  ft. ;  how  many  units  of  work  are  per- 
formed in  raising  this  ram  ?        Ans.   19040  foot-pounds. 

2.  How  many  units  of  work  are  required  to  raise  7  cwt. 
of  coal  from  a  mine  whose  depth  is  13  fathoms  ? 

Ans.  61152  foot-pounds, 

3.  A  horse  is  used  to  lift  the  earth  from  a  trench,  which 
he  does  by  means  of  a  cord  going  over  a  pulley.  He  pulls 
up,  twice  every  5  minutes,  a  man  weighing  130  lbs.,  and  a 
barrowf  ul  of  earth  weighing  260  lbs.  Each  time  the  horse 
goes  forward  40  ft. ;  find  the  units  of  work  done  by  the 
horse  per  hour.  Ans.  374400. 

4.  A  railway  train  of  T  tons  ascends  an  inclined  plane 
which  has  a  rise  of  e  ft.  in  100  ft.,  with  a  uniform  speed  of 
m  miles  per  hour ;  find  the  horse-power  of  the  engine,  the 
friction  beings  lbs.  per  ton. 

A       mT(p  +  22.4e)  __  _ 
Am'  —    375— J  H-R 

5.  A  railway  train  of  80  tons  ascends  an  incline  which 
rises  one  foot  in  50  ft.,  with  the  uniform  rate  of  15  miles 

*  One  ton  =  20  cwt.  =  2240  lbs. 


EXAMPLES.  419 

per  hour  ;  find  the  horse-power  of  the  engine,  the  friction 
being  8  lbs.  per  ton.  Am.   1G8.9G  H.-P. 

6.  If  a  horse  exert  a  traction  of  t  lbs.,  what  weight,  w 
will  he  pull  up  or  down  a  hill  of  small  inclination  which 
has  a  rise  of  e  in  100,  the  coefficient  being/? 

100* 

*«■ w  =  mr±e- 

7.  From  what  depth  will  an  engine  of  22  horse-power 
raise  13  tons  of  coal  in  an  hour  ?  Ans.  1496  ft. 

8.  An  engine  is  observed  to  raise  7  tons  of  material  an 
hour  from  a  mine  whose  depth  is  85  fathoms ;  find  the 
horse-power  of  the  engine,  supposing  -J  of  its  woik  to  be 
lost  in  transmission.  Ans.  4.8465  H.-P. 

9.  Eequired  the  horse-power  of  an  engine  that  would 
supply  a  city  with  water,  working  12  hours  a  day,  the 
water  to  be  raised  to  a  height  of  50  ft. ;  the  number  of 
inhabitants  being  130000,  and  each  person, to  use  5  gallons 
of  water  a  day,  the  gallon  weighing  8J  lbs.  nearly. 

Ans.  11.4  H.-P. 

10.  From  what  depth  will  an  engine  of  20  horse-power 
raise  600  cubic  feet  of  water  per  hour  ?     Ans.  1056  feet. 

11.  At  what  rate  per  hour  will  an  engine  of  30  horse- 
power draw  a  train  weighing  90  tons  gross,  the  resistance 
oeing  8  lbs.  per  ton  ?  Ans.  15.625  miles. 

12.  What  is  the  gross  weight  of  a  train  which  an  engine 
of  25  horse-power  will  draw  at  the  rate  of  25  miles  an 
hour,  resistances  being  8  lbs.  per  ton  ? 

Ans.  46.875  tons. 

13.  A  train  whose  gross  weight  is  80  tons  travels  at  the 
rate  of  20  miles  an  hour ;  if  the  resistance  is  8  lbs, 
per  ton,  what  is  the  horse-power  of  the  engine  ? 

Ans.   34v>FH.-P. 


420  EXAMPLES. 

14.  What  must  be  the  length  of  the  stroke  of  a  piston 
of  an  engine,  the  surface  of  which  is  1500  square  inches, 
which  makes  20  strokes  per  minute,  so  that  with  a  mean 
pressure  of  12  lbs.  on  eacli  square  inch  of  the  piston,  the 
engine  may  be  of  80  horse-power  ?  Ans.   7^  ft. 

15.  The  diameter  of  the  piston  of  an  engine  is  80  ins., 
the  length  of  the  stroke  is  10  ft.,  it  makes  11  strokes  per 
minute,  and  the  mean  pressure  of  the  steam  on  the  piston 
is  12  lbs.  per  square  inch  ;  what  is  the  horse-power  ? 

Ans.  201-06  H.-P. 

16.  The  cylinder  of  a  steam  engine  has  an  internal 
diameter  of  3  ft.,  the  length  of  the  stroke  is  6  ft.,  it  makes 
6  strokes  per  minute;  under  what  effective  pressure  per 
square  inch  would  it  have  to  work  in  order  that  75  horse- 
power may  be  done  on  the  piston  ?  Ans.  67-54  lbs. 

17.  It  is  said  that  a  horse,  walking  at  the  rate  of  2 J  miles 
an  hour,  can  do  1650000  units  of  work  in  an  hour ;  what 
force  in  pounds  does  he  continually  exert  ? 

Ans.   125  lbs. 

18.  Find  the  horse-power  of  an  engine  which  is  to  move 
at  the  rate  of  30  miles  an  hour,  the  weight  of  the  engine 
and  load  being  50  tons,  and  the  resistance  from  friction 
16  lbs.  per  ton.  Ans.  64  H.-P. 

19.  There  were  6000  cubic  ft.  of  water  in  a  mine  whose 
depth  is  60  fathoms,  when  an  engine  of  50  horse-power 
began  to  work  the  pump  ;  the  engine  continued  to  work  5 
hours  before  the  mine  was  cleared  of  the  water  ;  required 
the  number  of  cubic  ft.  of  water  which  had  run  into  the 
mine  during  the  5  hours,  supposing  J  of  the  work  of  the 
engine  to  be  lost  by  transmission.       Ans.  10500  cubic  ft. 

20.  Find  the  horse-power  of  a  steam  engine  which  will 
raise  30  cubic  ft.  of  water  per  minute  from  a  mine  440  ft, 
deep.  Ans.  25  H.-P. 


EXAMPLES.  421 

21.  If  a  pit  10  ft.  deep  with  an  area  of  4  square  ft.  be 
excavated  and  the  earth  thrown  up,  how  much  work  will 
have  been  done,  supposing  a  cubic  foot  of  earth  to  weigh 
90  lbs.  Ans.  18000  fi-lbs. 

22.  A  well-shaft  300  ft.  deep  and  5  ft.  in  diameter  is  full 
of  water:  how  many  units  of  work  must  be  expended  in 
getting  this  water  out  the  well ;  (L  e.,  irrespectively  of  any 
other  water  flowing  in)?  Ans.  55223202  ft.-lbs. 

23.  A  shaft  a  ft.  deep  is  full  of  water;  find  the  depth  of 
the  surface  of  the  water  when  one-quarter  of  the  work 
required  to  empty  the  shaft  has  been  done.  .        a 

J±71S.    —  it. 

A 

24.  The  diameter  of  the  cylinder  of  an  engine  is  80  ins., 
the  piston  makes  per  minute  8  strokes  of  10J  ft.  under  a 
mean  pressure  of  15  lbs.  per  square  inch  ;  the  modulus  of 
the  engine  is  0-55;  how  many  cubic  ft.  of  water  will  it 
raise  from  a  depth  of  112  ft.  in  one  minute  ? 

Ans.  485- 78  cub.  ft. 

25.  If  in  the  last  example  the  engine  raised  a  weight  of 
66433  lbs.  through  90  ft.  in  one  minute,  what  must  be  the 
mean  pressure  per  square  inch  on  the  piston  ? 

Ans.  26-37  lbs. 

26.  If  the  diameter  of  the  piston  of  the  engine  in  Ex.  24 
had  been  85  ins.,  what  addition  in  horse-power  would  that 
make  to  the  useful  power  of  the  engine  ? 

Ans.  13-28  H.-P. 

27.  If  an  engine  of  50  horse-power  raise  2860  cub.  ft.  of 
water  per  hour  from  a  mine  60  fathoms  deep,  find  the 
modulus  of  the  engine.  Ans.  -65. 

28.  Find  at  what  rate  an  engine  of  30  horse-power  could 
draw  a  train  weighing  50  tons  up  an  incline  of  1  in  280, 
the  resistance  from  friction  being  7  lbs.  per  ton. 

Ans.  1320  ft.  per  minute. 


422  EXAMPLES. 

29.  A  forge  hammer  weighing  300  lbs.  makes  100  lifts  a 
minute,  the  perpendicular  height  of  each  lift  being  2  ft.; 
what  is  the  horse-power  of  the  engine  that  gives  motion  to 
20  such  hammers ?  Ans.  36-36  H.-P. 

30.  An  engine  of  10  horse-power  raises  4000  lbs.  of  coal 
from  a  pit  1200  ft.  deep  in  an  hour,  and  also  gives  motion 
to  a  hammer  which  makes  50  lifts  in  a  minute,  each  lift 
having  a  perpendicular  height  of  4  ft.;  what  is  the  weight 
of  the  hammer  ?  Ans.  1250  lbs 

31.  Find  the  horse-power  of  the  engine  to  raise  T  tons  of 
coal  per  hour  from  a  pit  whose  depth  is  a  fathoms,  and  at 
the  same  time  to  give  motion  to  a  forge  hammer  weighing 
w  lbs.,  which  makes  n  lifts  per  minute  of  li  ft.  each. 

224ar  r  +  nhw  TT   _> 
AnS'  33006         H-R 

32.  Find  the  useful  work  done  by  a  fire  engine  per 
second  which  discharges  every  second  13  lbs.  of  water  with 
a  velocity  of  50  ft.  per  second.  Ans.  508  nearly. 

33.  A  railway  truck  weighs  m  tons ;  a  horse   draws  it 

along  horizontally,  the  resistance  being  n  lbs.  per  ton ;  in 

passing  over  a  space  s  the  velocity  changes  from  u  to  v ; 

find  the  work  done  by  the  horse  in  this  space. 

.        2240m  ,  ..   • 

Ans.  — - —  (v*  —  w1)  +  tnns. 

34.  The  weight  of  a  ram  is  600  lbs.,  and  at  the  end  of 
the  blow  has  a  velocity  of  32J  ft.;  what  work  has  been 
done  in  raising  it  ?  Ans.  9650. 

35.  Required  the  work  stored  in  a  cannon  ball  whose 
weight  is  32£  lbs.,  and  velocity  1500  ft.       Ans.   1125000. 

36.  A  ball,  weighing  20  lbs.,  is  projected  with  a  velocity 
of  60  ft.  a  second,  on  a  bowling-green  ;  what  space  will  the 
ball  move  over  before  it  comes  to  rest,  allowing  the  friction 
to  be  ^  the  weight  of  the  ball  ?  Ans.   1007-3  ft. 


EXAMPLES.  423 

37.  A  train,  weighing  193  tons,  has  a  velocity  of  30 
miles  an  hour  when  the  steam  is  turned  off;  how  far  will 
the  train  move  on  a  level  rail  before  coming  to  rest,  the 
1'riction  being  5-|  lbs.  per  ton  ?  Ans.   12256  ft. 

38.  A  train,  weighing  60  tons,  has  a  velocity  of  40  miles 
an  hour,  when  the  steam  is  turned  off,  how  far  will  it 
ascend  an  incline  of  1  in  100,  taking  friction  at  8  lbs.  a  ton  ? 

Ans.  3942J  ft. 

39.  A  carriage  of  1  ton  moves  on  a  level  rail  with  the 
speed  of  8  ft.  a  second;  through  what  space  must  the 
carriage  move  to  have  a  velocity  of  2  ft.,  supposing  friction 
to  be  6  lbs.  a  ton?  Ans.  348  ft. 

40.  If  the  carriage  in  the  last  example  moved  over  400 
feet  before  it  comes  to  a  state  of  rest,  what  is  the  resistance 
of  friction  per  ton?  Ans.  5.57  lbs. 

41.  A  force,  P,  acts  upon  a  body  parallel  to  the  plane ; 
find  the  space,  s,  moved  over  when  the  body  has  attained  a 
given  velocity,  v,  the  coefficient  of  friction  being/,  and  the 
body  weighing  w  lbs.  .  _  wv2 

AnS.     S    -      yyz  ~^~\' 

42.  Suppose  the  body  in  the  last  example  to  be  moved 
for  t  seconds  ;  required  (1)  the  velocity,  v,  acquired,  and 
(2)  the  work  stored. 

43.  A  body,  weighing  40  lbs.,  is  projected  along  a  rough 
horizontal  plane  with  a  velocity  of  150  ft.  per  second  ;  the 
coefficient  of  friction  is  ^ ;  find  the  work  done  against  fric- 
tion in  5  seconds.  Ans.  3500  foot-pounds. 

44.  A  body  weighing  50  lbs.,  is  projected  along  a  rough 
horizontal  plane  with  the  velocity  of  40  yards  per  second ; 
find  the  work  expended  when  the  body  comes  to  rest. 

Ans.  11250  ft.-lbs. 


424  .  EXAMPLES. 

45.  If  a  train  of  cars  weighing  100000  lbs.  is  moving  or, 
a  horizontal  track  with  a  velocity  of  40  miles  an  hour  when 
the  steam  is  turned  off;  through  what  space  will  it  move 
before  it  is  brought  to  rest  by  friction,  the  friction  being 
8  lbs.  per  ton  ?  A?is.  13374.8  ft. 

46.  What  amount  of  energy  is  acquired  by  a  body  weigh- 
ing 30  lbs.  that  falls  through  the  whole  length  of  a  rough 
inclined  plane,  the  height  of  which  is  30  ft.,  and  the  base 
100  ft.,  the  coefficient  of  friction  being  \  ? 

Ans.  300ft.-lbs. 

47.  If  a  train  of  cars,  weighing  T  tons,  ascend  an  incline 

having  a  raise  of  e  ft.  in  100  ft.,  with  the  velocity  v0  ft.  per 

second  when  the  steam  is  turned  off ;  through  what  space, 

x,  will  it  move  before  it  comes  to  a  state  of  rest,  the  friction 

beings  lbs.  per  ton  ?  .  1120v ft2 

Ans,  x  =  — — — ■— 5 -. 

g(22Ae  +  p) 

48.  Suppose  the  train,  in  Ex.  4,  Art.  217,  to  be  attached 
to  a  rope,  passing  round  a  wheel  at  the  top  of.  the  incline, 
which  has  an  empty  train  of  Tx  tons  attached  to  the  other 
extremity  of  the  rope;  what  velocity,  v,  will  the  train 
acquire  in  descending  s  ft.  of  the  incline  ? 


Ans-v  =  V^k~m 


1120 

49.  An  engine  of  35  horse-power  makes  20  revolutions 
per  minute,  the  weight  of  the  fly-wheel  is  20  tons  and  the 
diameter  is  20  ft.;  what  is  the  accumulated  energy  in  foot- 
pounds? Ans.  307054. 

50.  If  the  fly-wheel  in  the  last  example  lifted  a  weight  of 
4000  lbs.  through  3  ft.,  what  part  of  its  angular  velocity 
would  it  lose?  Ans.  -£T. 

51.  If  the  axis  of  the  above  fly-wheel  be  6  ins.  in 
diameter,  the  coefficient  of  friction  0-075,  what  fraction, 


EXAMPLES.  425 

approximately,  of  the  35  horse-power  is  expended  in  tinn- 
ing the  lly- wheel  ?  Ans.  ^. 

52.  In  pile  driving,  38  men  raised  a  ram  12  times  in  an 
hour;  the  weight  of  the  ram  was  12  cwt.,  and  the  height 
through  which  it  was  raised  140  ft;  find  the  work  done  by 
one  man  in  a  minute.  Ans.  990  ft.-lbs. 

53.  A  battering-ram,  weighing  2000  lbs.,  strikes  the 
head  of  a  pile  with  a  velocity  of  20  ft.  per  second ;  how  far 
will  it  drive  the  pile  if  the  constant  resistance  is  10000  lbs.? 

Ans.  1.25  ft. 

54.  A  nail  2  ins.  long  was  driven  into  a  block  by  suc- 
cessive blows  from  a  monkey  weighing  5.01  lbs.;  after  one 
blow  it  was  found  that  the  head  of  the  nail  projected  0-8 
of  an  inch  above  the  surface  of  the  block  ;  the  monkey  was 
then  raised  to  a  height  of  1.5  ft.,  and  allowed  to  fall  upon 
the  head  of  the  nail ;  after  this  blow  the  head  of  the  nail 
was  0.46  of  an  inch  above  the  surface;  find  the  force  which 
the  monkey  exerted  upon  the  head  of  the  nail  at  this  blow. 

Ans.  265.24  lbs. 

55.  The  monkey  of  a  pile-driver,  weighing  500  lbs.  is 
raised  to  a  height  of  20  ft.,  and  then  allowed  to  fall  upon 
the  head  of  a  pile,  which  is  driven  into  the  ground  1  inch 
by  the  blow;  find  the  force  which  the  monkey  exerted 
upon  the  head  of  the  pile.  Ans.  120000  lbs. 

56.  A  steam  hammer,  weighing  500  lbs.,  falls  through  a 

height  of  4  ft.  under  the  action  of  its  own  weight  and  a 

steam  pressure  of  1000  lbs.;    find  the  amount  of  work 

which  it  can  do  at  the  end  of  the  fall. 

Ans.  6000  ft.-lbs. 

57.  The  mean  section  of  a  stream  is  8  square  ft.;  its 
mean  velocity  is  40  ft.  per  minute;  it  has  a  fall  of  17|  ft.; 
it  is  required  to  raise  water  to  a  height  of  300  ft.  by  means 
of  a  water-wheel  whose  modulus  is  0.7  ;  how  many  cubic  ft 
will  it  raise  per  minute  ?  Ans.  13.07  cub.  ft. 


426  EXAMPLES. 

58.  To  what  height  would  the  wheel  in  the  last  example 
raise  2£  cub.  ft.  of  water  p^r  minute  ?         Ans.  1742f  ft. 

59.  The  mean  section  of  a  stream  is  1J  ft.  by  11  ft.;  its 
mean  velocity  is  2£  "miles  an  hour  ;  there  is  on  it  a  fall  of 
6  ft.  on  which  \s  erected  a  wheel  whose  modulus  is  0.?;  this 
wheel  is  employed  to  raise  the  hammers  of  a  forge,  each  of 
which  weighs  2  tons,  and  has  a  lift  of  l^ft.;  how  many 
lifts  of  a  hammer  will  the  wheel  yield  per  minute? 

Ans.  142  nearly. 

60.  In  the  last  example  determine  the  mean  depth  of 
the  stream  if  the  wheel  yields  135  lifts  per  minute. 

Ans.  1.43  ft. 

61.  In  Ex.  59,  how  many  cubic  ft.  of  water  must  descend 
the  fall  per  minute  to  yield  97  lifts  of  the  hammer  per 
minute  ?  Ans.  2483  cub.  ft. 

62.  A  stream  is  a  ft.  broad  and  b  ft.  deep,  and  flows  at 
the  rate  of  c  ft.  per  hour;  there  is  a  fall  of  h  ft. ;  the  water 
turns  a  machine  of  which  the  modulus  is  e  ;  find  the  num- 
ber of  bushel?  of  corn  which  the  machine  can  grind  in  an 
hour,  supposing  that  it  requires  m  units  of  work  per 
minute  for  one  hour  to  grind  a  bushel.       .       1000abc7ie 

AnS'  16  x  60m' 

63.  Down  a  14-ft.  fall  200  cub.  ft.  of  water  descend  every 
minute,  and  turn  a  wheel  whose  modulus  is  0.6.  The 
wheel  lifts  water  from  the  bottom  of  the  fall  to  a  height  of 
54  ft. ;  (1)  how  many  cubic  ft.  will  be  thus  raised  per 
minute?  (2)  If  the  water  were  raised  from  the  top  of  the 
fall  to  the  same  point,  what  would  the  number  of  cubic  ft. 
then  be?  Ans.   (1)  31.1  cub.  ft.;  (2)  34.7  cub.  ft. 

In  the  second  case  the  number  of  cub.  ft.  of  water  taken  from  the 
top  of  the  fall  being  %,  the  number  of  ft.  that  will  turn  the  wheel  will 
be  200  -  x. 

64.  Find  how  many  units  of  work  are  stored  up  in  a 


EXAMPLES.  427 

mill-pond  which  is  100  ft.  long,  50  ft.  broad,  and  3  ft.  deep, 
and  has  a  fall  of  8  ft.  Ans.  7500000. 

65.  There  are  three  distinct  levels  to  be  pumped  in  a 
mine,  the  first  100  fathoms  deep,  the  second  120,  the  third 
150  ;  30  cub.  ft.  of  water  are  to  come  from  the  first,  40  from 
the  second,  and  60  from  the  third  per  minute  ;  the  duty  of 
the  engine  is  70000000  for  a  bushel  of  coal.  Determine  (1) 
its  working  horse-power  and  (2)  the  consumption  of  coal 
per  hour.  "  Ans.   (1)  191  H.-R;  (2)  5.4  bushels. 

66.  In  the  last  example  suppose  there  is  another  level  of 
160  fathoms  to  be  pumped,  that  the  engine  does  as  much 
work  as  before  for  the  other  levels,  and  that  the  utmost 
power  of  the  engine  is  275  H.-P. ;  find  the  greatest  number 
of  cub.  ft.  of  water  that  can  be  raised  from  the  fourth  level. 

Ans.  46 J  cub.  ft. 

67.  A  variable  force  has  acted  through  8  ft.;  the  value 
of  the  force  taken  at  nine  successive  equidistant  points, 
including  the  first  and  the  last,  is  in  lbs.  10.204,  9.804, 
9.434,  9.090,  8.771,  8.475,  8.197,  7.937,  7.692;  find  the 
whole  work  done.  Ans.   70.641  foot-pounds. 

68.  The  value  of  a  variable  force,  taken  at  nine  succes- 
sive equidistant  points,  including  the  first  and  the  last 
points,  is  in  lbs.  2.4849,  2.5649,  2.6391,  2.7081,  2.7726, 
2.8332,  2.8904,  2.9444,  2.9957,  the  common  distance  between 
the  points  is  1  ft. ;  find  the  whole  work  done. 

Ans.  22.0957  foot-pounds. 

69.  A  train  whose  weight  is  100  tons  (including  the 
engine)  is  drawn  by  an  engine  of  150  horse-power,  the  fric- 
tion being  14  lbs.  per  ton,  and  all  other  resistances  neglect- 
ed ;  find  the  maximum  speed  which  the  engine  is  capable 
of  sustaining  on  a  level  rail.      Ans.  40^\  miles  per  hour. 

70.  If  the  train  described  in  the  last  example  be  moving 
at  a  particular  instant  with  a  velocity  of  15  miles  per  hour, 

% 


428  EXAMPLES. 

and  the  engine  working  at  full  power,  what  is  the  accelera- 
tion at  that  instant  ?     (Call  g  ==  32.)  ^?i5.  T4^-. 

71.  Find  the  horse-power  of  an  engine  required  to  drag  a 
train  of  100  tons  up  an  incline  of  1  in  50  with  a  velocity  of 
30  miles  an  hour,  the  friction  being  1400  lbs. 

Ans.  The  engine  must  be  of  not  less  than  470f  horse- 
power. This  is  somewhat  above  the  power  of  most  locomo- 
tive engines. 

72.  A  train,  of  200  tons  weight,  is  ascending  an  incline 
of  1  in  100  at  the  rate  of  30  miles  per  hour,  the  friction 
being  8  lbs.  per  ton.  The  steam  being  shut  off  and  the 
break  applied,  the  train  is  stopped  in  a  quarter  of  a  mile. 
Find  the  weight  of  the  break-van,  the  coefficient  of  fric- 
tion of  iron  on  iron  being  £.  Ans.  llT3f  tons. 


S 

* 


♦ 


CHAPTER    VI. 

MOMENT    OF    INERTIA.* 

224.  Moments  of  Inertia. — The  quantity  2wr  in 
which  m  is  the  mass  of  an  element  of  a  body,  and  r  its 
distance  from  an  axis,  occurs  frequently  in  problems  of 
rotation,  so  that  it  becomes  necessary  to  consider  it  in 
detail ;  it  is  called  the  moment  of  inertia  of  the  body  about 
the  axis  (Art.  218).  Hence,  "moment  of  inertia"  may  be 
denned  as  follows :  If  the  mass  of  every  particle  of  a  body  be 
multiplied  by  the  square  of  its  distance  from  a  straight  line, 
the  sum  of  the  products  so  formed  is  called  the  Moment  of 
Inertia  of  the  body  about  that  line. 

If  the  mass  of  every  particle  of  a  body  be  multiplied  by 
the  square  of  its  distance  from  a  given  plane  or  from  a 
given  point,  the  sum  of  the  products  so  formed  is  called  the 
moment  of  inertia  of  the  body  with  reference  to  that  plane 
or  that  point. 

If  the  body  be  referred  to  the  axes  of  x  and  y,  and  if  the 
mass  of  each  particle  be  multiplied  by  its  two  co-ordinates, 
x,  y,  the  sum  of  the  products  so  formed  is  called  the 
product  of  inertia  of  the  body  about  those  two  axes. 

If  dm  denote  the  mass  of  an  element,  p  its  distance  from 
the  axis,  and  /  the  moment  of  inertia,  we  have 

/  =  Zfdm.  (1) 

If  the  body  be  referred  to  rectangular  axes,  and  x,  y,  z, 
be  the  co-ordinates  of  any  element,  then,  according  to  the 
definitions,  the  moments  of  inertia  about  the  axes  of  x,  y, 
z,  respectively,  will  be 

*  This  term  was  introduced  by  Eider,  and  has  now  got  into  general  use  when- 
ever Rigid  Dynamics  is  studied. 


430  EXAMPLES. 

2  (y2  +  #)  dm,    S  (^  +  z2)  *w,    2  (z2  +  y2)  dm.     (2) 

The  moments  of  inertia  with  respect  to  the  planes  yz,  zx, 
xy  respectively,  are, 

2x2dm,     Zy2dm,     Z  z2dm.  (3) 

The  products  of  inertia  with  respect  to  the  axes  y  and  z, 
%  and  x,  x  and  y,  are 

£  yzdm,    X  zxdm,     2  zy^m.  (4) 

The  moment  of  inertia  with  respect  to  the  origin  is 

S  (z2  +  #2  +  22)  dim  =  2  rWw,  (5) 

where  r  is  the  distance  of  the  particle  from  the  origin. 

The  moment  of  inertia  of  a  lamina,  when  the  axis  lies  m 
it,  is  called  a  rectangular  moment  of  inertia,  and  when  it  is 
perpendicular  to  the  lamina  it  is  called  a  polar  moment  of 
inertia,  and  the  corresponding  axis  is  called  the  rectangular 
or  the  polar  axis. 

The  process  of  finding  moments  and  products  of  inertia 
is  merely  that  of  integration  ;  but  after  this  has  been  accom- 
plished for  the  simplest  axes  possible,  they  can  be  found 
without  integration  for  any  other  axes. 

EXAMPLES. 

1.  Find  the  moment  of  inertia  of  a  uniform  rod,  of  mass 
m,  and  length  I,  about  an  axis  through  its  centre  at  right 
angles  to  it. 

Let  x  be  the  distance  of  any  element  of  the  rod  from  the 
centre,  and  \i  the  mass  of  a  unit  of  length  ;  then  dm  =  \idx% 
which  in  (1)  gives  for  the  moment  of  inertia  2  \ix2dx,  or 


I  =    /    \ix2dx9 


EXAMPLE.  431 

remembering  that  the  symbol  of  summation,  £,  includes 
integration  in  the  cases  wherein  the  body  is  a  continuous 
mass. 
Hence  /  =  -f^P  =  -famK 

If  the  axis  be  drawn  through  one  end  of  the  rod  and 
perpendicular  to  its  length  we  shall  have  for  the  moment 
of  inertia 

/  =  \mK 

2.  Find  the  moment  of  inertia  of  a  rectangular  lamina* 
about  an  axis  through  its  centre,  parallel  to  one  of  its  sides. 

Let  b  and  d  denote  the  breadth  and  depth  respectively  of 
the  rectangle,  the  former  being  parallel  to  the  axis.  Im- 
agine the  lamina  composed  of  elementary  strips  of  length  b 
parallel  to  the  axis.  Let  the  distance  of  one  of  them  from 
the  axis  be  y,  and  its  breadth  dy  ;  then,  denoting  the  mass 
of  a  unit  of  area  by  //,  we  have  dm  =  pbdy,  which  in  (1) 
gives 

/\d 
\ibifdy  =  -^\ibdz  =  -&md?. 
\d 

If  the  axis  be  drawn  through  one  end  of  the  rectangle,  we 
shall  have  for  the  moment  of  inertia 

1  = 

3.  Find  the  moment  of  inertia  of  a  circular  lamina  with 
respect  to  an  axis  through  its  centre  and  perpendicular  to 
its  surface. 

Let  the  radius  =  a,  and  \i  the  mass  of  a  unit  of  area  as 
before,  then  we  have 


p2rt    pa 

=  V 

^0      *70 


2t   '       o  7    7/1        tiia*       ma* 
r3  dr  dd  =  -£—  =  — - 


*  In  all  cases  we  shall  assume  the  thickness  of  the  laminae  or  plates  to  b« 
Infinitesimal. 


432  FARALLEL   AXES. 

4.  Find  the  moment  of  inertia  of  a  circular  plate  (1) 
about  a  diameter  as  an  axis,  and  (2)  about  a  tangent. 

Ans.  (1)  l?na2;  (2)  \ma%. 

5.  Find  the  moment  of  inertia  of  a  square  plate,  (1) 
about  an  axis  through  its  centre  and  perpendicular  to  its 
plane,  (2)  about  an  axis  which  joins  the  middle  points  of 
two  opposite  sides,  and  (3)  about  an  axis  passing  through 
an  angular  point  of  the  plate,  and  perpendicular  to  its 
plane.  Let  a  =  the  side  of  the  plate  and  \i  the  mass  of  a 
unit  of  area. 

(2)  TV"«2;    (3)  \ma\ 

G.  Find  the  moment  of  inertia  of  an  isosceles  triangular 
plate,  (1)  about  an  axis  through  its  vertex  and  perpen- 
dicular to  its  plane,  and  (2)  about  an  axis  which  passes 
through  its  vertex  and  bisects  the  base. 

Let  2b  =  the  base  and  a  =  the  altitude,  then 

1=2  T  fa  ii  (a*  +  f)  dy  dx  =  ~  (3a*  +  P)  ;  (2)  \mb\ 

225.  Moments  of  Inertia  relative  to  Parallel 
Axes,  or  Planes. — The  moment  of  inertia  of  a  tody  about 
any  axis  is  equal  to  its  moment  of  inertia  about  a  parallel 
axis  through  the  centre  of  gravity  of  the  body,  phis  the 
product  of  the  mass  of  the  body  into  the  square  of  the  dis- 
tance betiveen  the  axes. 

Let  the  plane  of  the  paper  pass 
through  the  centre  of  gravity  of  the 
body,  and  be  perpendicular  to  the  two 
parallel  axes,  meeting  them  in  0  and 
G,  and  let  P  be  the  projection  of  any  c 
element   on   the   plane   of  the   paper. 


Fig. 'J  I 


EXAMPLES.  433 

Take  the  centre  of  gravity,  G,  as  origin,  the  fixed  axis 
through  it  perpendicular  to  the  plane  of  the  paper  as  the 
axis  of  z,  and  the  plane  through  this  and  the  parallel  axis 
for  that  of  zx ;  and  let  Ix  be  the  moment  of  inertia  about 
the  axis  through  G,  /  that  for  the  parallel  axis  through  0, 
a  the  distance,  OG,  between  the  axes,  and  (x,  y)  any  point, 
P.     Then  we  shall  have 

IA  =  2  (a*  +  if)  dm  ;  I  =  2  [(x  +  af  +  f]  dm. 

Hence        /  —  1^  =  2a  Sxdm  -f-  a2  Iidm  =  ahn, 

since  Zxdm  =  0,  as  the  centra  of  gravity  is  at  the  origin. 

...    l=  i^  +a*m,  (1) 

which  is  called  the  formula  of  reduction. 

Hence  the  moment  of  inertia  of  a  body  relative  to  any 
axis  can  be  found  when  that  for  the  parallel  axis  through 
its  centre  of  gravity  is  known. 

Cor.  1. — The  moments  of  inertia  of  a  body  are  the  same 
for  all  parallel  axes  situated  at  the  same  distance  from  its 
centre  of  gravity.  Also,  of  all  parallel  axes,  that  which 
passes  through  the  centre  of  gravity  of  a  body  has  the  least 
moment  of  inertia. 

Cor.  2. — It  is  evident  that  the  same  theorem  holds  if  the 
moments  of  inertia  be  taken  with  respect  to  parallel  planes, 
instead  of  parallel  axes. 

A  similar  property  also  connects  the  moment  of  inertia 
relative  to  any  point  with  that  relative  to  the  centre  of 
gravity  of  the  body. 

EXAMPLES. 

1.  The  moment  of  inertia  of  a  rectangle*  in  reference 
to  an  axis  through  its  centre  and  parallel  to  one  end  is 

*  See  Note  to  Ex.  2,  Art.  224  ;  strictly  Bpeaking,  an  area  has  a  moment  of  inertia 
no  more  than  it  has  weight. 


434  RADIUS   OF  GYRATIOX. 

T^mcP  ;  find  the  moment  of  inertia  in  reference  to  a  parallel 
axis  through  one  end. 
From  (1)  we  have 

d2 
I  ==  -j^md2  -f  -—  m  =  \md\ 

2.  The  moment  of  inertia  of  an  isosceles  triangle  about 
an  axis  through  its  vertex  and  perpendicular  to  its  plane 
is  \m  (3a2  +  b2),  (Art.  224,  Ex.  6) ;  find  its  moment  about 
a  parallel  axis  through  the  centre. 

From  (1)  we  have 

It  =  \m  (da2  +  ¥)  —  ia2m  =  \m  (\a2  -f  S2). 

3.  Find  the  moment  of  inertia  of  a  circle  about  an  axis 
through  its  circumference  and  perpendicular  to  its  plane 
(See  Ex.  3,  Art.  224).  Ans.  \ma\ 

4.  Find  the  moment  of  inertia  of  a  square  about  an  axis 
through  the  middle  point  of  one  of  its  sides  and  perpen- 
dicular to  its  plane  (Ex.  5,  Art.  224).  Ans.  -^ma2. 

226.  Radius  of  Gyration. — Let  k  be  such  a  quantity 
that  the  moment  of  inertia  =  mk2,  then  we  shall  have 

/  =  Zr2dm  =  ink2  (1) 

The  distance  k  is  called  the  radius  of  gyration  of  the 
body  with  respect  to  the  fixed  axis,  and  it  denotes  the 
distance  from  the  axis  to  that  point  into  which  if  the  whole 
mass  were  concentrated  the  moment  of  inertia  would  not  be 
altered.  The  point  into  which  the  body  might  be  concen- 
trated, without  altering  its  moment  of  inertia,  is  called  the 
centre  of  gyration.  When  the  fixed  axis  passes  through  the 
centre  of  gravity,  the  lenath  k  and  the  point  of  concentra- 
tion are  called  principal  radius  and  principal  centre  of 
gyration. 


RADIUS   OF  GYRATION.  455 

Lot  kl  =  the  principal  radius  of  gyration  and  rx  the 
distance  of  an  element  from  the  axis  through  the  centre  of 
gravity;  then  from  (1)  we  have 

mk2  =  £  r2dm 

=  S  rl2d»i  +  ?na2,  [by  (1)  of  Art.  225] 

=  mkt2  -f-  ma2', 

.-.     ^  =  ^2  + a2,  (2) 

from  which  it  appears  that  the  principal  radius  of  gyration 
is  the  least  radius  for  parallel  axes,  which  is  also  evident 
from  Cor.  1,  Art.  225. 

Sch. — In  homogeneous  bodies,  since  the  mass  of  any  part 
varies  directly  as  its  volume,  (1)  may  be  written 

SrtfF  =  Vk\  (3) 

where  d  V  denotes  the  element  of  volume,  and  V  the  entire 
volume  of  the  body. 

Hence,  in  homogeneous  bodies,  the  value  of  k  is  inde- 
pendent of  the  density  of  the  body,  and  depends  only  on  its 
form;  and  in  determining  the  moment  of  inertia,  we  may 
take  the  element  of  volume  or  weight  for  the  element  of 
mass,  and  the  total  volume  or  weight  of  the  body  instead 
of  its  mass. 

Also  in  finding  the  moment  of  inertia  of  a  lamina,  since 
k  is  independent  of  the  thickness  of  the  lamina,  we  may 
take  the  element  of  area  instead  of  the  element  of  mass, 
und  the  total  area  of  the  lamina  instead  of  its  mass. 

From  (1)  we  have 

k2  =  -•  (4) 

Similarly,  k2  =  ^,  (5) 


43 G  POLAB   MOMENT   OF  TXEL'TTA. 

hence,  the  square  of  the  radius  of  gyration  with  respect  to 
any  axis  equals  the  moment  of  inertia  with  respect  to  the 
same  axis  divided  by  the  mass. 

EXAMPLES. 

1.  Find  the  principal  radius  of  gyration  of  a  straight 
line. 

From  Ex.  1,  Art.  224,  we  have 

it  =  tV^2; 

therefore  from  (5)  we  have  hx2  =  -£%K 

2.  Find  the  principal  radius  of  gyration  of  a  circle  (1) 
with  respect  to  a  polar  axis,  and  (2)  with  respect  to  a 
rectangular  axis.  Ans.   (1)  |a8;  (2)  \a2. 

3.  Find  the  principal  radius  of  gyration  of  a  rectangle 
with  respect  to  a  rectangular  axis.  Ans.  -^d2. 

4.  Find  the  principal  radius  of  gyration  (1)  of  a  square 
with  respect  to  a  polar  axis,  and  (2)  of  an  isosceles  triangle 
with  respect  to  a  polar  axis. 

Ans.   (1)>2.    (2)  ±{%a2  +  b2). 

227.  Polar  Moment  of  Inertia. — If  any  thin  plate,  or 
lamina,  he  referred  to  two  rectangular  axes  and  x,  y  he  the 
co-ordinates  of  any  element,  then  (Art.  224)  the  moments 
of  inertia  about  the  axes  of  x  and  y  respectively,  are  S  y2dm 
and  2  x2dm  ;  and  therefore  the  moment  of  inertia  with 
respect  to  the  axis  drawn  perpendicular  to  the  plane  at  the 
intersection  of  the  axes  of  x  and  y  is 

2  (x2  +  y2)  dm. 

Hence  the  polar  mommt  of  inertia  of  any  lamina  is  equal 
to  the  sum  of  the  moments  of  inertia  with  respect  to  any  two 
rectangular  axes,  lying  in  the  plane  of  the  lamina. 


POLAR   MOMENT   OB1  INERTIA.  437 

Cor. — For  every  two  rectangular  axes  in  the  plane  of 
the  lamina,  at  any  point,  we  have 

2  x2dm  +  ^  \fdm  =  const. 

that  is,  the  sum  of  the  moments  of  inertia  with  respect  to  a 
pair  of  rectangular  axes  is  constant.  Hence,  if  one  be  a 
maximum,  the  other  is  a  minimum,  and  vice  versa. 

EXAMPLES. 

1.  Find  the  moment  of  inertia  of  a  rectangle  with  respect 
to  an  axis  through  its  centre  and  perpendicular  to  its  plane. 

From  Ex.  2,  Art.  224,  the  rectangular  moments  of 
inertia  are 

-^md2  and  -famb2 ; 

therefore  the  polar  moment  of  inertia  =  fem  (d?  +  b2) ; 
V  =  A  (d>  +  &)• 

2.  Find  the  moment  of  inertia  of  an  isosceles  triangle 
with  respect  to  an  axis  through  its  centre  parallel  to  its 
base,  a  being  the  altitude  and  and  2b  the  base. 

Ans.  -fama2;  fa2  =  -£%a2. 

228.  Moment  of  Inertia  of  a  Solid  of  Revolution, 
with  respect  to  its  Geometric  Axis. — Let  the  axis  be 
that  of  x;  and  let  the  equation  of  the  generating  curve 
be  y  —f(x).  Let  the  solid  be  divided  into  an  infinite 
number  of  circular  plates  perpendicular  to  the  axis  of 
revolution ;  let  the  density  be  uniform  and  \i  the  mass  of  a 
unit  of  volume  ;  and  denote  by  x  the  distance  of  the  centre 
of  any  circular  plate  from  the  origin,  y  its  radius,  and  dx 
its  thickness ;  then  the  moment  of  inertia  of  this  circular 
plate  about  an  axis  through  its  centre  and  perpendicular  to 
its  plane,  by  (Ex.  3,  Art.  224),  is 


438  EXAMPLES. 

therefore  the  moment  of  inertia  of  the  whole  solid  is 

f /[/(*)?<**;  (i) 

the  integration  being  taken  between  proper  limits. 

EXAMPLES. 

1.  Find  the  moment  of  inertia  of  a  right  circular  cone 
about  its  axis. 

Let  h  =  the  height  and  b  —  the  radius  of  the  base ; 

then   the   equation    of  the   generating  curve  is  y  =  j  x, 

wThich  in  (1)  gives  for  the  moment  of  inertia, 


=  jV^,  (since  m  =  -  f.ijib2)- 
Therefore    ht*  =  ^¥. 

2.  Find  the  moment  of  inertia  (1)  of  a  solid  cylinder 
about  its  axis,  b  being  its  radius  and  h  its  height,  and  (2) 
of  a  hollow  cylinder,  b  and  b'  being  the  external  and 
internal  radii.  Ana.   (1)  \mb2\  (2)  \m  {b2  +  b'2). 

3.  Find  the  moment  of  inertia  of  a  paraboloid  about  its 
axis,  h  being  its  altitude  and  b  the  radius  of  the  base. 

.         77  uhb* 

Ans.  -T-. 

229.  Moment  of  Inertia  of  a  Solid  of  Revolution, 
with  respect  to  an  Axis  Perpendicular  to  its  Geo- 
metric Axis. — Take  the  origin  at  the  intersection  of  the 


EXAMPLES.  439 

axis  of  revolution  with  the  axis  about  which  the  moment 
of  inertia  is  required ;  and  denoting  by  x  the  distance  of 
the  centre  of  any  circular  plate  from  the  origin,  y  its 
radius  and  dx  its  thickness,  we  have  for  the  moment  of 
inertia  of  this  circular  plate,  about  a  diameter,  by  Ex.  4, 
Art.  224, 

therefore  (Art.  225)  the  moment  of  inertia  of  this  plate 
about  the  parallel  axis  at  the  distance  x  from  it  is 

-~-  dx  -f  nfj,  qfx2  dx ; 

therefore  the  moment  of  inertia  of  the  whole  solid  is 

**/{£  +  **)**.  a) 

the  integration  being  taken  between  proper  limits. 

EXAMPLES. 

1.  Find  the  moment  of  inertia  of  a  right  circular  cone 
about  an  axis  through  its  vertex  and  perpendicular  to  its 
own  axis. 

Let  h  =  the  height  and  b  =  the  radius  of  the  base,  then 
the  moment  of  inertia  from  (1) 

=  ftm  (47*2  +  V). 

2.  Find  the  moment  of  inertia  of  a  cone,  whose  altitude 
=  h,  and  the  radius  of  whose  base  =  b,  about  an  axis 
through  its  centre  of  gravity  and  perpendicular  to  its  own 
axis.  Ans.  -^m  (h2  +  4J2). 


440  EXAMPLES. 

3.   Find  the  moment  of  inertia  of  a  paraboloid  of  revolt 

tion  about  an  axis  through  its  vertex  and  perpendicular  to 

its  own  axis,  the  altitude  being  //  and  the  radius  of  the 

base  b.  .       -nuhb2  /7„       070, 

Am.  -!~-~  (b2  +  U2). 

230.  Moment  of  Inertia  of  Various  Solid  Bodies. 

EXAM  PLE  S. 

1.  Find  the  moment  of  inertia  of  a  rectangular  parallel- 
epiped about  an  axis  through  its  centre  of  gravity  and  par- 
allel to  an  edge. 

Let  the  edges  be  a,  b,  c;  since  a  parallelopiped  may  be 
conceived  as  consisting  of  an  infinite  number  of  rectangular 
laminae,  each  of  which  has  the  same  radius  of  gyration 
relative  to  an  axis  perpendicular  to  its  plane,  it  follows 
that  the  radius  of  gyration  of  the  parallelopiped  is  the 
same  as  that  of  the  laminss.  Hence,  the  moments  of 
inertia  relative  to  three  axes  through  the  centre  and  par- 
allel to  the  edges  a,  b,  c,  respectively,  are  by  Ex.  1,  Art. 
227,  Tyw  (b2  +  c2),  &m  {a2  +  c2),  &m  (a2  +  fa). 

2.  Find  the  moment  of  inertia  of  a  rectangular  parallel- 
opiped about  an  edge. 

This  may  be  obtained  immediately  from  the  last  exam- 
ple by  using  Art.  225,  or  otherwise  independently  as 
follows : 

Take  the  three  edges  a,  b,  c  for  the  axes  of  x,  y,  z, 
respectively  ;  let  \i  be  the  mass  of  a  unit  of  volume,  then 
the  moment  of  inertia  relative  to  the  edge  a  is 

pa     r*b     r*c 

=    /     /     /  Y>  (f  +  *)  dx  dy  dz 

"0    «'0    t70 


MOTION   OF  INERTIA    OF  A    LAMINA, 


441 


and  similarly  for  the  moments  of  inertia,  about  the  edges 
b  and  c. 

The  moment  of  inertia  of  a  cube  whose  edge  is  a  with 
respect  to  one  of  its  edges  is  |//«5  =  \ma%. 

3.  Find  the  moment  of  inertia  of  a  segment  of  a  sphere 
relative  to  a  diameter  parallel  to  the  plane  of  section,  the 
radius  of  the  sphere  being  a  and  the  distance  of  the  plane 
section  from  the  centre  b. 

Ans.  -^rr/i  (lGrt5  +  IdaW  -f  I0a2b*  —  9b5). 

231.  Moment  of  Inertia  of  a  Lamina  with  respect 
to  any  Axis. — When  the  moment  of  inertia  of  a  plane 
figure  about  any  axis  is  known,  wre  easily  find  the  moment 
of  inertia  about  any  parallel  axis  (Art.  225) ;  also,  when 
the  moments  of  inertia  about  two  rectangular  axes  in  the 
plane  of  the  figure  are  known,  the  moment  of  inertia  about 
the  straight  line  at  right  angles  to  the  plane  of  these  axes 
at  their  intersection  is  known  immediately,  (Art.  227)  ;  we 
now  proceed  to  find  the  moment  of  inertia  about  any 
straight  line  in  the  plane  inclined  to  these  axes  at  any 
angle. 

Through  any  point,  0,  as 
origin,  draw  two  rectangular 
axes,  OX,  OY,  in  the  plane  of 
the  lamina;  and  draw  any 
straight  line,  Ox,  in  the  plane. 
It  is  required  to  find  the  mo-  Fig- 

ment   of   inertia  about  Ox    in 
terms  of  the  moments  of  inertia  about  OX  and  OY. 

Let  P  be  any  point  of  the  lamina,  x,  y,  its  rectangular, 
and  r,  6,  its  polar  co-ordinates,  p  =  PM,  and  a  the  angle 
rcOX.  Then  if  I  be  the  moment  of  inertia  of  the  lamina 
relative  to  Ox,  a  and  b  the  moments  of  inertia  relative  to 
the  axes  of  x  and  y  respectively,  and  h  the  product  of 
inertia  relative  to  the  same  axes,  we  have 


442  PRINCIPAL   AXES   OF  A   BODY, 

1=1,  p2dm  =  Ir2  sin2  (0  —  a)  dm 
=  2  (y  cos  a  —  x  sin  a)2  dm 

=  cos2  a  S  ?/2dw  +  sin2  a  2  #2^m  —  2  sin  «  cos  a  2  #?/e?m 
=  #  cos2  «  -J-  b  sin2  «  —  2A  sin  a  cos  «.  (1) 

If  we  choose  the  axes  so  that  the  term  h  or  2  xydm  =  0, 
the  expression  for  I  becomes  much  simpler.  The  pair  of 
axes  so  chosen  are  called  the  principal  axes  at  the  point ; 
and  the  corresponding  moments  of  inertia  are  called  the 
principal  moments  of  iyiertia  of  the  lamina,  relative  to  the 
point. 

If  A  and  B  represent  these  principal  moments  of  inertia, 
(1)  becomes 

I  =  A  cos2  a  +  B  sin2  a.  (2) 

Hence,  the  moment  of  inertia  of  a  lamina  with  respect  to 
any  axis  through  a  point  may  be  found  when  the  principal 
moments  with  respect  to  the  point  are  determined. 

232.  Principal  Axes  of  a  Body. — At  any  point  of  a 
rigid  body  and  in  any  plane  there  is  a  pair  of  principal 
axes. 

Let  OX,  OY  (Fig.  92),  be  any  rectangular  axes  in  the 
plane;  let  Ox,  Oy,  be  another  set  of  rectangular  axes  in 
the  same  plane,  inclined  to  the  former  at  an  angle  « ;  let 
a,  b,  and  h,  as  before,  denote  the  moments  and  product  of 
inertia  about  OX,  OY,  and  let  (x1,  y')  be  any  point,  P, 
referred  to  the  axes  Ox,  Oy.  Then,  using  the  notation  of 
the  last  article,  we  have 

xf  =  r  cos  (6  —  a) ;    y'  =  r  sin  (0  —  a) ; 

X  x'y'dm  =  |2  r2  sin  2  (0  —  a)  dm 

=  cos  2«  £  r2  sin  0  cos  0  dm 

—  J  sin  2«  2  r2  (cos2  0  —  sin2  0)  dm. 

Putting  this  =  0,  and  solving  for  «,  we  obtain 


THREE    PRINCIPAL   AXES. 


443 


tan  2a  =  *& 


22  r"1  sin  8  cos  0  dm 


2  r*  (cos2  0  -  sin2  6)  dm 

22  xy  dm  2h 

-  2  (a;2  —  if)  dm  ~  V^'a 


(1) 


As  the  tangent  of  an  angle  may  have  any  value,  positive 
or  negative,  from  0  to  oo ,  it  follows  that  (1)  will  always 
give  a  real  value  for  2a,  so  there  is  always  a  set  of  princi- 
pal axes  ;  that  is,  at  every  point  in  a  body  there  exists  one 
pair  of  rectangular  axes  for  which  the  quantity  h  or 
2  xy  dm  =  0. 

Cor. — It  may  also  be  shown  that  at  every  point  of  a 
rigid  body  there  are  three  axes  at  right  angles  to  one 
another,  for  which  the  products  of  inertia  vanish.* 

*  Let  a,  ft,  c,  be  the  moments  of  inertia  about  three 
axes,  OX,  OY,  OZ,  at  right  angles  to  one  another  ;  d,  «, 
/,  the  products  of  inertia  (2  myz,  2  mzx,  2  mxyy  re- 
spectively). Let  Ox  be  any  line  drawn  through  the 
origin,  making  angles  a,  /?,  y,  with  the  co-ordinate 
axes. 

Let  OL,  LM,  MP,  be  the  coordinates  x,  y,  z,  of  any 
point  P  of  the  body  at  which  an  element  of  mass  m  is 
situated.    Draw  PN  perpendicular  to  Ox. 

Projecting  the  broken  line,  OLMP,  on  ON,  (Art. 
102),  we  have 

ON  =  x  cos  a  +  y  cos  /?  +  z  cos  y  ; 

also  OP2  =  x*  +  y2  +  z2,    and    1  =  cosa  a  +  cos2  /?  +  cos3  y. 

The  moment  of  inertia  I  about  Ox  =  SmPNJ 

=  2m  (OP2  -  ON2) 

=  2m  [x2  +  y*  +  z2  —  (a  cos  a  +  y  cos  /?  +  z  cos  y)2] 

=  2m[(a:,,  +  y2  +  z2)(cos2  a  +  cos2 /?  +  cos2  y)—(x  cos  a  +  y  cos  /?  +  z  cos  y)«] 

=  Tm(y*  +  z*)  cos2  a  +  2m  (z2  +  re2)  cos2  /?  +  2m  («2  +  y2)  cos'  y 

—  22myz  cos  /?  cos  y  —  22mzx-  cos  y  cos  a  -  22m  cos  a  cos  /? 
=  a  cos2  a  +  b  cos2  /?  +  c  cos2  y  —  2d  cos  /?  cos  y 

-  2«  COS  y  COS  a  -  2/  COS  a  cos  /?.  (1) 

To  represent  this  geometrically,  take  a  point  Q  on  ON  ;  and  Jet  its  distance 
from  O  be  /•,  and  its  co-ordinates  be  a\,  y,,  zx.    Then 

x%  =  r  cos  a,    y^.  —  r  cos  /?,    2X  =  r  cos  y. 


Fig.9ltf 


444  THREE  PRINCIPAL   AXES. 

Sch. — In  many  cases  the  position  of  the  principal  axea 
can  be  seen  at  once.  Suppose,  for  example,  we  wish  the 
principal  axis  for  a  rectangle  when  the  given  point  is  the 
centre.  Draw  through  the  centre  straight  lines  parallel  to 
the  sides  of  the  rectangle  ;  then  these  will  be  the  principal 

Therefore  (1)  becomes 

T      axS  +  byx*  +  czS  -  2dy,z,  -  2ez1xx  -  2/3^ 

I  = p; •  MP 

But  the  equation 

axS  +  byS  +  czS  -  Myxzx  -  2ezxxx  -  %fxxyx  =  1,  (3) 

denotes  an  ellipsoid  whose  centre  is  at  O ;  because  a,  b,  c  are  necessarily  positive, 
since  a  moment  of  inertia  is  essentially  positive,  being  the  sum  of  a  number  of 
squares.    If  theD  Q  is  a  point  on  this  ellipsoid,  (2)  becomes 

1  =  ZmPN2  =  —-% 

or  the  moment  of  inertia  about  any  line  through  O,  is  measured  by  the  square  of 
the  reciprocal  of  the  radius  vector  of  this  ellipsoid,  which  coincides  with  the 
line. 

This  is  called  the  momenta!  ellipsoid,  and  was  first  used  by  Cauchy,  Exercises  de 
Math.,  Vol.  II.  It  has  no  physical  existence,  but  is  an  artifice  to  bring  under  the 
methods  of  geometry  the  properties  of  moments  of  inertia.  The  momental  ellip- 
soid has  a  definite  form  for  every  point  of  a  rigid  body. 

Now  every  ellipsoid  has  three  axes,  to  which  if  it  is  referred,  the  coefficients  of 
yz,  zx,  xy  vanish,  and   therefore  (3),  when  transformed  to  these  axes  takes  the 

form 

Axt*  +  ByS  +  CzS  =  1;  (4) 

and  hence  (1)  or  (2)  when  referred  to  these  axes,  becomes 

I  =  A  COS8  a  +  B  cos2  0  +  C  cosa  y,  (5) 

where  A.  B,  C,  are  the  moments  of  inertia  of  the  body  about  these  axes. 

When  three  rectangular  axes,  meeting  in  a  given  point,  are  chosen  so  that  the 
products  of  inertia  all  vanish,  they  are  called  the  principal  axes  at  the  given 
point. 

The  three  planes  through  any  two  principal  axes  are  called  the  principal  planes 
at  the  given  point. 

The  moments  of  inertia  about  the  principal  axes  at  any  point  are  called  iheprin* 
cipal  moments  of  inertia  at  that  point. 

If  the  three  principal  moments  of  inertia  of  a  body  are  equal  to  one  another,  the 
ellipsoid  (4)  becomes  a  sphere,  since  A  =  B  =  C ;  and  therefore  the  moment  of 
inertia  about  every  other  axis  is  equal  to  these,  for  (5)  becomes 

I  =  A  (cos3  a  +  cos8  0  +  cos2  >)  =  A ; 

and  every  axis  is  a  principal  axis.  (See  Routh's  Rigid  Dynamics,  p.  12,  Price's 
Anal.  Mech's,  Vol.  II,  p.  156,  Pirie's  Rigid  Dynamics,  p.  76,  etc.) 


THREE  PRINCIPAL   AXES.  445 

axes;  because  for  every  element,  dm,  on  one  side  of  the 
axis  of  x  at  the  point  (x,  y),  there  is  another  element  of 
equal  mass  on  the  other  side  at  the  point  (x,  — y).  Hence, 
2  xy  dm  consists  of  terms  which  may  be  arranged  in  pairs, 
so  that  the  two  terms  in  a  pair  arc  numerically  equal  but 
of  opposite  signs  ;  and  therefore  2  xy  dm  —  0. 

Again,  if  in  any  uniform  body  a  straight  line  can  be 
drawn  with  respect  to  which  the  body  is  exactly  symmetri- 
cal, this  must  be  a  principal  axis  at  every  point  in  its 
length.  Any  diameter  of  a  uniform  circle  or  sphere  or  the 
axis  of  a  parabola  or  ellipse  or  hyperbola  is  a  principal  axis 
at  any  point  in  its  line;  but  the  diagonal  of  a  rectangular 
plate  is  not  for  this  reason  a  principal  axis  at  its  middle 
point,  for  every  straight  line  drawn  perpendicular  to  it  is 
not  equally  divided  by  it. 

Let  the  body  be  symmetrical  about  the  plane  of  xy,  then 
for  every  element  dm,  on  one  side  of  the  plane  at  the  point 
(x,  y,  z),  there  is  another  element  of  equal  mass  on  the 
other  side  at  the  point  (x,  y,  — z).  Hence,  for  such  a  body 
2  xz  dm  =  0  and  2  yz  dm  =  0.  If  the  body  be  a  lamina  in 
the  plane  of  xy,  then  z  of  every  element  is  zero,  and  we 
have  again  2  xz  dm  =  0,  2  yz  dm  =  0. 

Thus,  in  the  case  of  the  ellipsoid,  the  three  principal 
sections  are  all  planes  of  symmetry,  and  therefore  the  three 
axes  of  the  ellipsoid  are  principal  axes.  Also,  at  every 
point  in  a  lamina  one  principal  axis  is  the  perpendicular  to 
the  plane  of  the  lamina. 


examples. 

1.  Find  the  moment  of  inertia  of  a  rectangular  lamina 
about  a  diagonal. 

From  Ex.  2,  Art.  224,  the  moments  of  inertia  about  two 
lines  through  the  centre  parallel  to  the  sides  (principa' 
moments  of  inertia)  are 


446  EXAMPLES. 

•femd2    and    -j^mb?; 

where  b  and  d  are  the  breadth  and  depth  respectively. 

Also,  if  «  be  the  angle  which  the  diagonal  makes  with 
the  side  b,  we  have 

sin2  a  =  „ Tn »     cos2  « 


£2  +  #>       —    "   —   £2  +  d2 

Substituting  these  values  for  A,  B,  sin2  a,  cos2  «,  in  (2)  of 
Art.  231,  we  have 

b2  d2 

I  =  ^md2-——-  +  ^?nb2 


=  i»* 


£2  +  ^2     ■      12—    £2  +  ^2 


#  +  d2 


2.  Find  the  moment  of  inertia  of  an  isosceles  triangular 
plate  about  an  axis  through  its  centre  and  inclined  at  an 
angle  a  to  its  axis  of  symmetry,  a  being  its  altitude  and  2b 
its  base.  Ans.  \m  (\a2  cos2  a  +  b2  sin2  «). 

3.  Find  the  moment  of  inertia  of  a  square  plate  about  a 
diagonal,  a  being  a  side  of  the  square.  Ans.  -^ma2. 

233.  Products  of  Inertia. — The  value  of  the  product 
of  inertia  at  any  point  may  be  made  to  depend  on  the  value 
of  the  product  of  inertia  for  parallel  axes  through  the  cen- 
tre of  gravity.  Let  (x,  y)  be  the  position  of  any  element, 
dm,  referred  to  axes  through  any  assigned  point ;  (x\  y')  the 
position  of  the  element  referred  to  parallel  axes  through 
the  centre  of  gravity,  and  (h,  k)  the  centre  of  gravity 
referred  to  the  first  pair  of  axes.     Then 

x  =  x'  +  li,     y  =  y'  +  k; 

therefore      Zxydm  =  Z  (x'  +  h)  (y'  +  k)  dm 

=  2  x'y'  dm  +  hk  2  dm,  (1) 

0,  and  £  my'  =  0. 


EXAMPLES.  447 

Sch. — By  (1)  we  may  often  find  the  product  of  inertia 
for  an  assigned  origin  and  axes.  Thus,  suppose  we  require 
the  product  of  inertia  in  the  case  of  a  rectangle,  when  the 
origin  is  at  the  corner,  and  the  axes  are  the  edges  which 
meet  at  that  corner.  By  Art.  232,  Sch.  we  have  Zx'y'dm 
=  0;  therefore  from  (1)  we  have 

TLxy&m  z=z  hhZdm ; 

and  as  h  and  h  are  known,  being  half  the  lengths  of  the 
edges  of  the  rectangle  to  which  they  are  respectively 
parallel,  the  product  of  inertia  is  known. 

EXAMPLES. 

Find  the  expressions  for  the  moments  of  inertia  in  the 
following,  the  bodies  being  supposed  homogeneous  in  all 
cases. 

1.  The  moment  of  inertia  of  a  rod  of  length  a,  with 
respect  to  an  axis  perpendicular  to  the  rod  and  at  a  distance 

(5  +  4 

2.  The  moment  of  inertia  of  an  arc  of  a  circle  whose 
radius  is  a  and  which  subtends  an  angle  2«  at  the  centre,  (1) 
about  an  axis  through  its  centre  perpendicular  to  its  plane, 
(2)  about  an  axis  through  its  middle  point  perpendicular  to 
its  plane,  (3)  about  the  diameter  which  bisects  the  arc. 

sin  2«\  a2 
/2* 

3.  The  moment  of  inertia  of  the  arc  of  a  complete 
cycloid  whose  length  is  a  with  respect  to  its  base. 

Arts.  -j^ma2. 

4  The  moment  of  inertia  of  an  equilateral  triangle,  of 
side  a,  relative  to  a  line  in  its  plane,  parallel  to  a  side,  at 
the  distance  d  from  its  centre  of  gravity. 

Ans.  m  I—  -f  d2)> 


d  from  its  middle  point. 

Ans.  m 


i        f,^        o     /^x  ^     /-,      sin  a\    ,     /ft.       L      sin  2 
Ans.  (I)  ma2-,  (2)  2m  1 1 J  a2;  (3)  m  f  1 x — \ 


448  EXAMPLES. 

5.  Given  a  triangle  whose  sides  are  a,  b,  c,  and  whose 
perpendiculars  on  these  sides,  from  the  opposite  vertices 
arep,  q,  r,  respectively;  find  the  moment  of  inertia  of  the 
triangle  about  a  line  drawn  through  each  vertex  and 
parallel  respectively,  (1)  to  the  side  a,  (2)  to  the  side  b,  (3) 
to  the  side  c.  Ans.   (1)  ^ni]j2;  (2)  %?nq2 ;  (3)  %mr2. 

6.  Find  the  moment  of  inertia  of  the  triangle  in  the  last 
example  relative  to  the  three  lines  drawn  through  the 
centre  of  gravity  of  the  triangle  and  parallel  respectively 
to  the  sides  a,  b,  c.  Ans.  -^mp2 ;  jgMq2 ;  -^mr2. 

7.  Find  the  moment  of  inertia  of  the  triangle  in  Ex.  5, 
relative  to  the  three  sides  a,  b,  c,  respectively. 

Ans.  \mj)2\  jtnq2;  \mr2. 

8.  The  moment  of  inertia  of  a  right  angled  triangle,  of 
hypothenuse  c,  relative  to  a  perpendicular  to  its  plane 
passing  through  the  right  angle.  Ans.  \m&. 

9.  The  moment  of  inertia  of  a  ring  whose  outer  and 
inner  radii  are  a  and  b  respectively,  (1)  with  respect  to  a 
polar  axis  through  its  centre,  and  (2)  with  respect  to  a 
diameter.  Ans.   (1)  \m  (a2  +  b2)  ;  (2)  \m  [a2  +  ^). 

10.  The  moment  of  inertia  of  an  ellipse,  (1)  with  respect 
to  its  major  axis,  (2)  with  respect  to  its  minor  axis,  and  (3) 
with  respect  to  an  axis  through  its  centre  and  perpendicular 
to  its  plane. 

Ans.    (1)  lmb2\    (2)  Ima2;  <3)  \m  (a2  +  b2). 

11.  The  moment  of  inertia  of  the  surface  of  a  sphere  of 
radius  a  about  its  diameter.  Ans.  ^ma2. 

12.  The  moment  of  inertia  of  a  right  prism  whose  base 
is  a  right  angled  triangle,  with  respect  to  an  axis  passing 
through  the  centres  of  gravity  of  the  ends,  the  sides  con- 
taining the  right  angle  of  the  triangular  base  being  a  and  i 
and  the  height  of  the  prism  c.  Ans.  -fern  (a2  -\-b2). 


EXAMPLES.  449 

13.  The  moment  of  inertia  of  a  right  prism  whose  height 
Is  c,  about  an  axis  passing  through  the  centres  of  gravity  of 
the  ends,  the  base  of  the  prism  being  an  isosceles  triangle 
whose  base  is  a  and  height  b.  .        ,  /a2       &\ 

AnS'  *U   +3T"- 

14.  The  moment  of  inertia  of  a  sphere  of  radius  a,  (1) 
relative  to  a  diameter,  and  (2)  relative  to  a  tangent. 

Arts.   (1)  \ma?\  (2) \ma2. 

15.  The  moment  of  inertia,  about  its  axis  of  rotation,  (1) 
of  a  prolate  spheroid,  and  (2)  of  an  oblate  spheroid. 

Ans.  (1)  \m&\  (2)  %ma2. 

16.  The  moment  of  inertia  of  a  cylinder,  relative  to  an 
axis  perpendicular  to  its  own  axis  and  intersecting  it,  (1)  at 
a  distance  c  from  its  end,  (2)  at  the  end  of  the  axis,  and  (3) 
at  the  middle  point  of  the  axis,  the  altitude  of  the  cylinder 
being  h  and  radius  of  its  base  a. 

Ans.   (1)  \ma2  +  $m  (h2  —  3hc  +  3c2) ; 

(2)  -&m  (3a2  +  Ah2) ;  (3)  ^m  (A2  +  3a2). 

17.  The  moment  of  inertia  of  an  ellipse  about  a  central 
radius  vector  r,  making  an  angle  a  with  the  major-axis. 

Ans.  ±m  — —  • 
4      r2 

18.  The  moment  of  inertia  of  the  area  of  a  parabola  cut 
off  by  any  ordinate  at  a  distance  x,  from  the  vertex,  (1) 
about  the  tangent  at  the  vertex,  and  (2)  about  the  axis  of 
the  parabola. 

Ans.  fynx2  ;  (2)  \my2  where  y  is  the  ordinate  correspond- 
ing to  x, 

J9.  The  moment  of  inertia  of  the  area  of  the  lemniscate, 
r3  =  a2  cos  20,  about  a  line  through  the  origin  in  its  plane 
and  perpendicular  to  its  axis.  Ans.  -fam  (3n-.-f  8)  a2. 


450  EXAMPLES. 

20.  The  moment  of  inertia  of  the  ellipsoid, 

a*  +  0>  +  *  -  L 

about  the  axis  «,  J,  £,  respectively. 

^ws.  (1)  |m  (ga  +  c2) ;  (2)  |w  (c*  +  a*) 
(3)  \m  (a2  +  d2). 


CHAPTER    VII. 

ROTATORY    MOTION. 

234.  Impressed  and  Effective  Forces. — All  forces 
acting  on  a  body  other  than  the  mutual  actions  of  the 
particles,  are  called  the  Impressed  Forces  that  act  on  the 
body. 

Thus,  when  a  ball  is  thrown  in  vacuo,  the  impressed 
force  is  gravity ;  if  a  ball  is  rotating  about  a  vertical  axis, 
the  impressed  forces  are  gravity  and  the  reaction  of  the 
axis. 

The  impressed  or  external  forces  are  the  cause  of  the  motion  and  of 
all  the  other  forces.  Which  are  the  impressed  forces  depends  upon 
the  particular  system  which  is  under  consideration.  The  same  force 
may  be  external  to  one  system  and  internal  to  another.  Thus,  the 
pressure  between  the  foot  of  a  man  and  the  deck  of  a  ship  on  which 
he  is,  is  external  to  the  ship  and  also  to  the  man  and  is  the  cause  of 
Ms  own  forward  motion  and  of  a  slight  backward  motion  of  the  ship ; 
but  if  the  man  and  ship  are  considered  as  parts  of  one  system  the 
pressure  is  internal. 

When  a  particle  is  moving  as  part  of  a  rigid  body,  it  is 
acted  on  by  the  external  impressed  forces  and  also  by  the 
molecular  reactions  of  the  other  particles.  Now  if  this 
particle  were  considered  as  separated  from  the  rest  of  the 
body,  and  all  the  forces  removed,  there  is  some  one  force 
which,  singly,  would  move  it  in  the  same  way  as  before. 
This  force  is  called  the  Effective  Force  on  the  particle ;  it  is 
evidently  the  resultant  of  the  impressed  and  molecular 
forces  on  the  particle. 

Thus,  the  effective  force  is  that  part  of  the  impressed  force  which 
is  effective  in  causing  actual  motion.  It  is  the  force  which  is  required 
for  producing  the  deviation  from  the  straight  line  and  the  change  of 


452  D'ALEMBERT'S  PRINCIPLE. 

velocity.  If  a  particle  is  revolving  with  constant  velocity  rouiwf  a 
fixed  axis,  the  effective  force  is  the  centripetal  force  (Art.  198).  if  a 
heavy  body  falls  without  rotation,  the  whole  force  of  gravity  is 
effective  ;  but  if  it  is  rotating  about  a  horizontal  axis  the  weight  goes 
partly  to  balance  the  pressure  on  the  axis. 

If  we  suppose  the  particle  of  mass  m  to  be  at  the  point 
(x,  y,  z)  at  any  time,  t,  and  resolve  the  forces  acting  on  it 
into  the  three  axial  components,  X,  Y,  Z,  the  motion  may 
be  found  [Art.  168  (2)]  by  solving  the  simultaneous  equa- 
tions 

(fix        ~         d2y        T,         ePz        „  ,^ 

If  we  regard  a  rigid  body  as  one  in  which  the  particles 
retain  invariable  positions  with  respect  to  one  another,  so 
that  no  external  force  can  alter  them  (Art.  43),  we  might 
write  down  the  equations  of  the  several  particles  in  accord- 
ance with  (1),  if  all  the  forces  were  known.  Such,  how- 
ever, is  not  the  case.  We  know  nothing  of  the  mutual 
actions  of  the  particles,  and  consequently  cannot  determine 
the  motion  of  the  body  by  calculating  the  motion  of  its 
particles  separately.  When  there  are  several  rigid  bodies 
which  mutually  act  and  react  on  one  another  the  problem 
becomes  still  more  complicated. 

235.  D'Alembert's  Principle.*— By  D'Alembert's 
Principle,  however,  all  the  necessary  equations  may  be 
obtained  without  writing  down  the  equations  of  motion  of 
the  several  particles,  and  without  any  assumption  as  to  the 
nature  of  the  mutual  actions  except  the  following,  which 
may  be  regarded  as  a  natural  consequence  of  the  laws  of 
motion. 

The  internal  actions  and  reactions  of  any  system  of  rigid 
bodies  in  motion  are  in  equilibrium  among  themselves. 

*  Introduced  by  D'Alembert  in  1743. 


D*ALEMBEI?T'S   PRINCIPLE.  453 

The  axial  accelerations  of  the  particle  of  mass  m3  which 

(Pec      (ffill      (fiz 

is   moving   as   part   of  a  rigid    body,   are    — ,   -~,  — . 

Cll         (it         Clt 

Let/  be  their  resultant,  then  the  effective  force  is  measured 
by  mf  Let  i^and  R  be  the  resultants  of  the  impressed  and 
molecular  forces,  respectively,  on  the  particle.  Then  mf 
is  the  resultant  of  i^and  R.  Hence  if  mf  be  reversed,  the 
three  forces,  F,  R,  and  mf  are  in  equilibrium. 

The  same  reasoning  may  be  applied  to  every  particle  of 
each  body  of  the  system,  thus  furnishing  three  groups  of 
forces,  similar,  respectively,  to  F,  R,  and  mf;  and  these 
three  groups  will  form  a  system  of  forces  in  equilibrium. 
Now  by  D'Alembert's  principle  the  group  R  will  itself 
form  a  system  of  forces  in  equilibrium.  Whence  it  follows 
that  the  group  F  will  be  in  equilibrium  with  the  group  mf 
Hence, 

If  forces  equal  and  exactly  opposite  to  the  effective  forces 
were  applied  at  each  particle  of  the  system,  they  ivould  be 
in  equilibrium  with  the  impressed  forces. 

That  is,  D* Alember? s  principle  asserts  that  the  whole 
effective  forces  of  a  system  are  together  equivalent  to  the 
imp  ressed  forces, 

Sch. — By  this  principle  the  solution  of  a  problem  in 
Kinetics  is  reduced  to  a  problem  in  Statics  as  follows :  We 
first  choose  the  co-ordinates  by  means  of  which  the  position 
of  the  system  in  space  may  be  fixed.  We  then  express  the 
effective  forces  on  each  element  in  terms  of  its  co-ordinates. 
These  effective  forces,  reversed,  will  be  in  equilibrium 
with  the  given  impressed  forces.  Lastly,  the  equations  of 
motion  for  each  body  may  be  formed,  as  is  usually  done  in 
Statics,  by  resolving  in  three  directions  and  taking  mo- 
ments about  three  straight  lines.  (See  Eouth's  Rigid 
Dynamics,  Pirie's  "Rigid  Dynamics,  Pratt's  Mech's,  Price's 
Anal.  Mech's,  Vol.  II.) 


454  ROTATION  OF  A   RIGID   BODY. 

236.  Rotation  of  a   Rigid   Body  about   a   Fixed 
Axis   under  the   Action   of   any   Forces.— Let    any 

plane  passing  through  the  axis  of  rotation  and  fixed  in 
space  be  taken  as  a  plane  of  reference.  Let  m  be  the  mass 
of  any  element  of  the  body,  r  its  distance  from  the  axis, 
and  6  the  angle  which  a  plane  through  the  axis  and  the 
element  makes  with  the  plane  of  reference. 

Then  the  velocity  of  m  in  a  direction  perpendicular  to 

in 

the   plane   containing  the   element  and   the  axis  is  r -n' 

ax 

The   moment   of  the   momentum*  of   this   particle   about 

a       •   .       2de 

the  axis  is  trur-^* 
dt 

all  the  particles  is 


the  axis  is  mr2-^--*    Hence  the  moment  of  the  momenta  of 
dt 


W§.  (1) 

Since  the  particles  of  the  body  are  rigidly  connected, 

rid 

it  is  clear  that  -y  is  the  same  for  every  particle,  and  is  the 

angular  velocity  of  the  body.      Hence  the  moment  of  the 

momenta  of  all  the  particles  of  the  body  about  the  axis  is  the 

moment  of  inertia  of  the  body  about  the  axis  multiplied  by 

the  angular  velocity. 

The  acceleration  of  m  perpendicular  to  the  direction  in 

d?Q 
which  r  is  measured  is  r-s,  and  therefore  the  moment  of 

cPO 
the  moving  forces  of  m  about  the  axis  is  mr2-j-»    Hence, 

the  moment  of  the  effective  forces  of  all  the  particles  of  the 
body  about  the  axis  is 

2^2— ,  (2) 

which  is  the  moment  of  inertia  of  the  body  about  the  axit 
multiplied  by  the  angular  acceleration. 

*  Called  also  Angular  Momentum.    (See  Pirie's  Rigid  Dynamics,  p.  44,) 


ROTATION  OF  A   RIGID   BODY.  455 

(1)  Let  the  forces  be  impulsive  (Art.  202)  ;  let  w,  w',  be 
the  angular  velocities  just  before  and  just  after  the  action 
of  the  forces,  and  N  the  moment  of  the  impressed  forces 
about  the  axis  of  rotation,  by  which  the  motion  is  pro- 
duced. 

Then,  since  by  D'Alembert's  principle  the  effective 
forces  when  reversed  are  in  equilibrium  with  the  impressed 
forces,  we  have  from  (1) 

c/  2  mr2  -wS  mr2  =  N\ 

if 
2  mr2 


moment  of  impulse  about  axis# 
~  moment  of  inertia    about  axis' 


(3) 


that  is,  the  change  in  the  angular  velocity  of  a  body,  pro- 
duced by  an  impulse,  is  equal  to  the  moment  of  the  impulse 
divided  by  the  moment  of  inertia  of  the  body. 

(2)  Let  the   forces  be   finite.      Then   taking  moments 
about  the  axis  as  before,  we  have  from  (2) 


m      n 


dt2       S  mr2 

moment  of  forces  about  axis# 
moment  of  inertia  about  axis ' 


(*) 


that  is,  the  angular  acceleration  of  a  body,  produced  by  a 
force,  is  equal  to  the  moment  of  the  force  divided  by  the 
moment  of  inertia  of  the  body. 

By  integrating  (4)  we  shall  know  the  angle  through 
which  the  body  has  revolved  in  a  given  time.  Two  arbi- 
trary constants  will  appear  in  the  integrations,  whose 
values  are  to  be  determined  from  the  given  initial  values 

rift 

of  0  and  ~    Thus  the  whole  motion  can  be  found,  and 
dt 


456  EXAMPLE. 

we  shall  consequently  be  able  to  determine  the  position  oi 
the  body  at  any  instant. 

Sch. — It  appears  from  (3)  and  (4)  that  the  motion 
of  a  rigid  body  round  a  fixed  axis,  under  the  action  of  any 
forces,  depends  on  (1)  the  moment  of  the  forces  about 
that  axis,  and  (2)  the  moment  of  inertia  of  the  body  about 
the  axis.  If  the  whole  mass  of  the  body  were  concentrated 
into  its  centre  of  gyration  (Art.  226),  and  attached  to  the 
fixed  axis  of  rotation  by  a  rod  without  mass,  whose  length 
is  the  radius  of  gyration,  and  if  this  system  were  acted  on 
by  forces  having  the  same  moment  as  before,  and  were  set 
in  motion  with  the  same  initial  values  of  6  and  the  angular 
velocity,  then  the  whole  subsequent  angular  motion  of  the 
rod  would  be  the  same  as  that  of  the  body.  Hence,  we  may 
say  briefly,  that  a  body  turning  about  a  fixed  axis  is 
kinetically  given  when  its  mass  and  radius  of  gyration  are 
known. 

EXAM  PLE. 

A  rough  circular  horizontal  board  is  capable  of  revolving 
freely  round  a  vertical  axis  through  its  centre.  A  man 
walks  on  and  round  at  the  edge  of  the  board;  when  he 
has  completed  the  circuit  what   will  be  his  position  in 


Let  a  be  the  radius  of  the  board,  M  and  M'  the  masses 
of  the  board  and  man  respectively  ;  6  and  6'  the  angles 
described  by  the  board  and  man,  and  i^the  action  between 
the  feet  of  .the  man  and  the  board. 

The  equation  of  motion  of  the  board  by  (4)  is 

Since  the  action  between  the  man  and  the  board  is  con- 
tinually tangent  to  the  path  described  by  the  man,  the 
equation  of  motion  of  the  man  is,  by  (5)  of  Art.  20, 


TEE  COMPOUND   PENDULUM, 

cm' 


45? 


M'a 


dt* 


Eliminating  .Fand  integrating  twice,  the^constant  being 
/sero  in  both  cases,  because  the  man  and  board  start  from 
rest,  we  get 

Mkfd  =  M'aW.  (1) 

When  the  man  has  completed  the  circuit  we  have  0  -f  0' 

a2 
=  2tt;  also  &,2  =  -•     Substituting  these  in  (1)  we  get 


6'  = 


2rrM 


2M'  +  M' 


which  gives  the  angle  in  space  described  by  the  man. 
If  M  =  31' 9  this  becomes 


and 


e  =  i-tt, 


which  is  the  angle  in  space  described  by  the  board.     (See 
Roiith?s  Rigid  Dynamics,  p.  67.) 

237.  The  Compound  Pendulum. — A  body  moves  about 
a  fixed  horizontal  axis  acted  on  by  gravity  only,  to  determine 
the  motion. 

Let  ABO  be  a  section  of  the  body  made  by 
the  plane  of  the  paper  passing  through  G, 
the  centre  of  gravity,  and  cutting  the  axis 
of  rotation  perpendicularly  at  0.  Let  6  = 
the  angle  which  OG  makes  with  the  vertical 
OY;  and  let  h  =  OG,  &,  =  the  principal 
radius  of  gyration,  and  M  =  the  mass  of 
the  body.  "Then  by  (4)  of  Art.  236,  we 
have 

20 


£58  THE   COMPOUND   PENDULUM. 

dffl  _  Mgh  sin  0  __     _  Mgh  sin 6 
.    dt2  ~        Zm?*     ==  ,         ~W<? 

=  -  wh* sin  °  [by  (2)  of  Art-  m^>  w 

the  negative  sign  being  taken  because  0  is  a  decreasing 
function  of  the  time. 

This  equation  cannot  be  integrated  in  finite  terms,  but 
if  the  oscillations  be  small,  we  may  develop  sin  0  and  reject 
all  powers  above  the  first,  and  (1)  will  become 

Multiplying  by  2  d6  and  integrating,  and  supposing  that 

the  body  began  to  move   when  0   was  equal  to   <*,  (2) 
becomes 

Hence  denoting  the  time  of  a  complete  oscillation  by  T, 
we  have 

which  gives  the  time  in  seconds,  when  h  and  &,  are  meas- 
ured in  feet  and  g  =  32.18. 

When  a  heavy  body  vibrates  about  a  horizontal  axis,  by 
the  force  of  gravity,  it  is  called  a  compound  pendulum. 

Cor.  1. — If  we  suppose  the  whole  mass  of  the  compound 
pendulum  to  be  concentrated  into  a  single  point,  and  this 
point  connected  with  the  axis  by  a  medium  without  weight, 
it  becomes  a  simple  pendulum  (Art.  194).  Denoting  the 
distance  of  the  point  of  concentration  from  the  axis  by  I, 
we  have  for  the  time  of  an  oscillation,  by  (1)  of  Art.  194, 


CENTRES   OF  OSCILLATION  AND   SUSPENSION.       459 

n  a  I  —  If  the  point  be  so  chosen  that  the  simple  pendu- 
lum will  perform  an  oscillation  in  the  same  time  as  the 
compound  pendulum,  these  two  expressions  for  the  time  of 
an  oscillation  must  be  equal  to  each  other,  and  we  shall 
have 


J  = 


h 


«  h  +  *?  =  00',  (4) 

(0'  being  the  point  of  concentration). 

Cor.  2. — This  length  is  called  the  length  of  the  simph 
equivalent  pendulum ;  the  point  0  is  called  the  centre  of 
suspension*;  the  point  0',  into  which  the  mass  of  the  com- 
pound pendulum  must  be  concentrated  so  that  it  will 
oscillate  in  the  same  time  as  before,  is  called  the  centre  of 
oscillation;  and  a  line  through  the  centre  of  oscillation 
and  parallel  to  the  axis  of  suspension  is  called  an  axis  of 
oscillation. 

From  (4)  we  have 

•      (l-7i)h  =  ht*; 


or 


GO'.  GO  =  h*.  (5) 


Now  (5)  would  not  be  altered  if  the  place  of  0  and  0' 
were  interchanged ;  hence  if  0'  be  made  the  centre  of 
suspension,  then  0  will  be  the  centre  of  oscillation.  '  Thus 
the  centres  of  oscillation  and  of  suspension  are  convertible, 
and  the  time  of  oscillation  about  each  is  the  same. 

Cor.  3. — Putting  the  derivative  of  I  with  respect  to  h  in 
(4)  equal  to  sero,  and  solving  for  h,  we  get 


460  EXAMPLES. 

which  makes  I  a  minimum,  and  therefore  makes  t  a  mmi- 
mum.  Hence,  when  the  axis  of  suspension  passes  through 
the  principal  centre  of  gyration  the  time  of  oscillation  is  a 
mini  mum. 

Rem. — The  problem  of  determining  the  law  under  which  a  heavy 
body  swings  about  a  horizontal  axis  is  one  of  the  most  important  in 
the  history  of  science.  A  simple  pendulum  is  a  thing  of  theory  ;  our 
accurate  knowledge  of  the  acceleration  of  gravity  depends  therefore 
on  our  understanding  the  rigid  or  compound  pendulum.  This  was 
the  first  problem  to  wThich  D'Alembert  applied  his  principle. 

The  problem  was  called  in  the  days  of  D'Alembert,  the  "  centre  of 
oscillation."  It  was  required  to  find  if  there  were  a  point  at  which 
the  whole  mass  of  the  body  might  be  concentrated,  so  as  to  form  a 
simple  pendulum  whose  law  of  oscillation  was  the  same. 

The  position  of  the  centre  of  oscillation  of  a  body  was  first  correctly 
determined  by  Huyghens  and  published  at  Paris  in  1673.  As 
D'Alembert's  principle  was  not  known  at  that  time,  Huyghens  had  to 
discover  some  principle  for  himself.* 


EXAMPLES. 

1.  A  material  straight  line  oscillates  about  an  axis  per- 
pendicular to  its  length;  find  the  length  of  the  equivalent 
simple  pendulum. 

Let  'Za  =  the  length  of  the  line,  and  h  the  distance  of  its 
centre  of  gravity  from  the  point  of  suspension.     Then  since 


&J2  =  — ,  we  have  from  (4) 


i-i  +  g.  a) 


Cor.  1. — If  the  point  of  suspension  be  at  the  extremity 
of  the  line  (1)  becomes 


*  Routh's  Rigid  Dynamics,  p.  69. 


EXAMPLES.  4G1 

that  is,  the  length  of  the  equivalent  simple  pendulum  is 
two-thirds  of  the  length  of  the  rod. 

Cor.  2. — Let  h  =  %a ;  then  (1)  becomes 

I  =  •£«. 

Hence,  the  time  of  an  oscillation  is  the  same,  whether  the 
line  be  suspended  from  one  extremity,  or  from  a  point  one- 
third  of  its  length  from  the  extremity.  This  also  illustrates 
the  convertibility  of  the  centres  of  oscillation  and  of  sus- 
pension (See  Cor.  2). 

Cor.  3. — If  h  =  10a,  then  (1)  becomes 

I  =  *fra. 

2.  A  circular  arc  oscillates  about  an  axis  through  its 
middle  point  perpendicular  to  the  plane  of  the  arc.  Prove 
that  the  length  of  the  simple  equivalent  pendulum  is 
independent  of  the  length  of  the  arc,  and  is  equal  to  twice 
the  radius. 

From  Ex.  2,  Art.  233,  we  have 

k*  =  h*  +  ^  =  2(1  -S^)a*. 
From  Ex.  1,  Art.  78,  we  have 


sin  a 

h  =  a  —  a « 


Therefore  (4)  becomes 

.        .  „  L        sin  «\           /.        sin  «\        _ 
I  =  2az[  1 J  -j-  a  ( 1 1  =  2a. 


462  LENGTH   OF  THE  SECOND'S  PENDULUM. 

3.  A  right  cone  oscillates  about  an  axis  passing  through 
its  vertex  and  perpendicular  to  its  own  axis  ;  it  is  required 
to  find  the  length  of  the  simple  equivalent  pendulum,  (1) 
when  h  is  the  altitude  of  the  cone  and  b  the  radius  of  the 
base,  and  (2)  when  the  altitude  =  the  radius  of  the  base  =  h. 

An*.  (l)^gP;  (2)  h. 

That  is,  in  the  second  cone,  the  centre  of  oscillation  is  in 
the  centre  of  the  base ;  so  that  the  times  of  oscillation  are 
equal  for  axes  through  the  vertex  and  the  centre  of  the 
base  perpendicular  to  the  axis  of  the  cone. 

4.  A  sphere,  radius  a,  oscillates  about  an  axis  ;  find  the 
length  of  the  simple  equivalent  pendulum,  (1)  when  the 
axis  is  tangent  to  the  sphere,  (2)  when  it  is  distant  10tf 
from  the  centre  of  the  sphere,  and  (3)  when  it  is  distant 

-  from  the  centre  of  the  sphere. 

Ans.  (I)  %a;  (2)  *fra;  (3)-^. 

238.    The    Length    of    the     Second's    Pendulum 

Determined  Experimentally. — The  time  of  oscillation 

h  2 
of  a  compound  pendulum  depends  on  h  -f-  -=r-  by  (4)  of 

Art.  237.  But  there  are  difficulties  in  the  way  of  determin- 
ing h  and  Tcv  The  centre,  G,  can  not  be  got  at,  and,  as 
every  body  is  more  or  less  irregular  and  variable  in  density, 
k1  cannot  be  calculated  with  sufficient  accuracy.  These 
quantities  must  therefore  be  determined  from  experiments. 
Bessel  observed  the  times  of  oscillation  about  different 
axes,  the  distances  between  which  were  very  accurately 
known.  Captain  Kater  employed  the  property  of  the 
convertibility  of  the  centres  of  suspension  and  oscillation 
(Art.  237,  Cor.  2),  as  follows : 

Let  the  pendulum  consist  of  an  ordinary  straight  bar,  CO,  and  a 
small  weight,  m,  which  may  be  clamped  to  it  by  means  of  a  screw, 
and  shifted  from  one  position  to  another  on  the  pendulum.     At  the 


VT]m 


Q 


LENGTH  OF  THE  SECOND'S  TENBULtlM.  4(33 

poirts   C   and   0   in    two  triangular  aper-  rj  /^-> v 

tures,  at  the  distance  I  apart,  let  two  knife      C^^^  (      J 

edges  of  hard  steel   bo  placed  parallel  to 

each  other,  and    at    right  angles    to  the 

pendulum,  so  that  it  may  vibrate  on  either 

of  them,  as  in  Fig.  94.     Let  m  be  shifted 

till  it  is  found  that  the  times  of  oscillation 

about  0  and  0  are  exactly  the  same.     It         /^~*\ 

remains  only  to  measure  CO,  and  observe         v^/ 

the  time  of  oscillation.     The  distance  be-  Fig.,94 

tween  the  two  points  C  and  0  is  the  length 

of  the  simple  equivalent  pendulum.     This  distance  between  the  knife 

edges  was  measured  by  Captain  Kater  with  the  greatest  care.     The 

mean  of  three  measurements  differed  by  less  than  a  ten-thousandth 

of  an  inch  from  each  of  the  separate  measurements. 

The  time  of  a  single  vibration  cannot  be  observed  directly,  because 
this  would  require  the  fraction  of  a  second  of  time  as  shown  by  the 
clock,  to  be  estimated  either  by  the  eye  or  ear.  The  difficulty  may 
be  overcome  by  observing  the  time,  say  of  a  thousand  vibrations,  and 
thus  the  error  of  the  time  of  a  single  vibration  is  divided  by  a 
thousand.  The  labor  of  so  much  counting  may  however  be  avoided 
by  the  use  of  "  the  method  of  coincidences."  The  pendulum  is  placed 
in  front  of  a  clock  pendulum  whose  time  of  vibration  is  slightly 
different.  Certain  marks  made  on  the  two  pendulums  are  observed 
by  a  telescope  at  the  lowest  point  of  their  arcs  of  vibration.  The  field 
of  view  is  limited  by  a  diaphragm  to  a  narrow  aperture  across  which 
the  marks  are  seen  to  pass.  At  each  succeeding  vibration  one 
pendulum  follows  the  other  more  closely,  and  at  last  its  mark  is 
completely  covered  by  the  other  during  their  passage  across  the  field 
of  view  of  the  telescope.  After  a  few  vibrations  it  appears  again 
preceding  the  other.  In  the  interval  from  one  disappearance  to  the 
next,  one  pendulum  has  made,  as  nearly  as  possible,  one  complete 
oscillation  more  than  the  other.  In  this  manner  530  half-vibrations  of 
a  clock  pendulum,  each  equal  to  a  second,  were  found  to  correspond  to 
532  of  Captain  Kater's  pendulum.  The  ratio  of  the  times  of  vibra- 
tion of  the  pendulum  and  the  clock  pendulum  may  thus  be  calculated 
with  extreme  accuracy.  The  rate  of  going  of  the  clock  must  then  be 
found  by  astronomical  means. 

The  time  of  vibration  thus  found  will  require  several  corrections 
which  are  called  "reductions."  For  instance,  if  the  oscillation  be 
not  so  small  that  we  can  put  sin  0  =  6  in  Art.  237,  we  must  make  a 
reduction  to  infinitely  small  arcs.     Another  reduction  is  necessary  if 


1G4        MOTION  OF  A    BODY   WHEN    UNCONSTRAINED. 

vre  wish  to  reduce  the  result  to  what  it  would  have  been  at  the  lev<eJ 
of  the  sea.  The  attraction  of  the  intervening  land  may  be  allowed 
for  by  Dr.  Young's  rule,  (Phil.  Trans.,  1819).  We  may  thus  obtain 
the  force  of  gravity  at  the  level  of  the  sea,  supposing  all  the  land 
9-Love  this  level  were  cut  off  and  the  sea  constrained  to  keep  its 
oresent  level.  As  the  level  of  the  sea  is  altered  by  the  attraction  of 
ohe  land,  further  corrections  are  still  necessary  if  we  wish  to  reduce 
the  result  to  the  surface  of  that  spheroid  which  most  nearly  repre- 
sents the  earth.  See  Routh's  Rigid  Dynamics,  p.  77.  For  the  details 
of  this  experiment  the  student  is  referred  to  the  Phil.  Trans,  for  1818, 
and  to  Vol.  X. 

239.  Motion  of  a  Body  "when  Unconstrained. — If 

an  impulse  be  communicated  to  any  point  of  a  free  body 
in  a  direction  not  passing  through  the  centre  of  gravity,  \t 
will  produce  both  translation  and  rotation. 

Let  P  be  the  impulse  imparted  to 
the  body  at  A.  At  B,  on  the  opposite 
side  of  the  centre  G,  a  distance  GB 
=  AG,  let  two  opposite  impulses  be 
applied,  each  equal  to  \P  \  they  will 
not  alter  the  effect.  Now  if  \P 
applied  at  A  is  combined  with  the  \P  '9' 

at  B  which  acts  in  the  same  direction,  their  resultant  is  P, 
acting  at  G  and  in  the  same  direction,- and  this  produces 
translation  only.  The  remaining  \P  at  A  combined  with 
the  remaining  \P  at  B,  which  acts  in  the  opposite  direc- 
tion, form  a  couple  which  produces  rotation  about  the 
centre  G. 

Hence,  when  a  tody  receives  an  impulse  in  a  direction 
which  does  not  pass  through  the  centre  of  gravity,  that  centre 
will  assume  a  motion  of  translation  as  though  the  impulse 
were  applied  immediately  to  it;  and  the  body  will  have  a 
motion  of  rotation  about  the  centre  of  gravity,  as  though 
that  point  were  fixed. 

240.  Centre  of  Percussion. — Axis  of  Spontaneous 
Rotation. — Let  Mv  represent  the  impulse  impressed  upon 


ip 


CENTRE   OF  PERCUSSION.  465 

the  body  (Fig.  90)  whose  mass  is  M9  and 
It  the  perpendicular  distance,  GO,  from 
the  centre  of  gravity,  G,  to  the  line  of 
action,  OF,  of  the  impulse.  The  centre 
of  gravity  will  assume  a  motion  of  trans- 
lation with  the  velocity  v,  in  a  direction 
parallel  to  that  of  the  impulsive  force. 
Then  from  (3)  of  Art.  236,  we  have  for  the  angulai 
velocity 

Mvh        vh 


G) 


Mk?  ~  k? 


The  absolute  velocity  of  each  point  of  the  body  will  be 
compounded  of  the  twro  velocities  of  translation  and  rota- 
tion. The  point  0,  for  example,  to  which  the  impulse  is 
applied,  has  a  velocity  of  translation,  Oa,  equal  to  that  of 
the  centre  of  gravity,  and  a  velocity  of  rotation,  ab,  about 
the  centre  of  gravity;  so  that  the  velocity  of  any  point  at 
a  distance  a  from  the  centre,  G,  will  be  expressed  by 
v  ±  au ;  the  upper  or  lower  sign  being  taken  according  as 
the  point  is,  or  is  not,  on  the  same  side  of  the  centre  of 
gravity  as  the  point  0.  Thus,  if  we  consider  the  motion  of 
the  body  for  a  very  short  interval  of  time,  the  line  OGC 
will  assume  the  position  bG'G,  the  point  C  remaining  at 
rest  during  this  interval ;  that  is,  while  the  point  C  would 
be  carried  forward  over  the  line  Cc  by  the  motion  of  trans- 
lation, it  would  be  carried  backward  through  the  same 
distance  by  the  motion  of  rotation.  Hence,  since  the  abso- 
lute velocity  of  G  is  zero,  we  have 

v  —  ao)  =  0 ; 
and  hence  denoting  OC  by  I  we  have 


466  AXIS   OF  SPONTANEOUS  ROTATION. 

r=  h  +  ^.  (*, 

Now  if  there  bad  been  a  fixed  axis  through  0  perpen- 
dicular  to  the  plane  of  motion,  the  initial  motion  would 
have  been  precisely  the  same,  and  this  fixed  axis  evidently 
would  not  have  received  any  pressure  from  the  impulse. 

When  a  rigid  body  rotates  about  a  fixed  axis,  and  the 
body  can  be  so  struck  that  there  is  no  pressure  on  the  axis, 
any  point  in  the  line  of  action  of  the  force  is  called  a  centre 
of  percussion. 

When  the  line  of  action  of  the  blow  is  given  and  the 
body  is  free  from  all  constraint,  so  that  it  is  capable  of 
translation  as  well  as  of  rotation,  the  axis  about  which  the 
body  begins  to  turn  is  called  the  axis  of  spontaneous  rota- 
tion. It  obviously  coincides  with  the  position  of  the  fixed 
axis  in  the  first  case. 

Cok.  1. — From  (1)  we  have 

ah  =  GO-GO  =  h?; 

hence  the  points  0  and  C  are  convertible,  that  is,  if  the 
axis  of  rotation  he  supposed  to  pass  through  the  point  0, 
the  centre  of  spontaneous  rotation  will  coincide  with  the  cen- 
tre of  percussion. 

Coe.  2. — From  (2)  it  follows,  by  comparison  with  (4)  of 
Art.  237,  that  if  the  axis  of  spontaneous  rotation  coincides 
with  the  axis  of  suspension,  the  centre  of  percussion  coin- 
cides with  the  centre  of  oscillation.. 

Sch. — It  is  evident  that  if  there  be  a  fixed  obstacle  at  0, 
and  it  be  struck  by  the  body  00  rotating  about  a  fixed 
axis  through  C,  the  obstacle  will  receive  the  whole  force 
of  the  moving  body,  and  the  axis  will  not  receive  any, 
Heuce  the  centre  of  percussion  also  determines  the  position 
in  which  a  fixed  obstacle  must  be  placed,  on  which  if  the 
rotating  body  impinges  and  is  brought  to  rest,  the  axis  of 
rotation  will  suffer  no  pressure. 


EXAMPLES.  46? 

An  axis  through  the  centre  of  gravity,  parallel  to  the 
axis  of  spontaneous  rotation,  is  called  the  axis  of  instantane- 
ous rotation.     A  free  body  rotates  about  this  axis  (Art.  239). 

EXAMPLES. 

1.  Find  the  centre  of  percussion  of  a  circular  plate  of 

radius  a  capable  of  rotating  about  an  axis  which  touches  it. 

a2 
Here  k?  =  j,  and  h  —  a.     Hence  from  (2)  we  have 

1  ~  a  +  4  =  ia' 

2.  A  cylinder  is  capable  of  rotating  about  the  diameter 
of  one  of  its  circular  ends  ;  find  the  centre  of  percussion. 
Let  a  =  its  length,  and  b  =  the  radius  of  its  base. 

3b2  +  4«2 


Ans.  I  = 


(ja 


Hence  if  3b2  =  2a2,  the  centre  of  percussion  will  be  at 
the  end  of  the  cylinder.  If  b  is  very  small  compared  with 
a,  ?==|«;  thus  if  a  straight  rod  of  small  transverse  section 
is  held  by  one  end  in  the  hand,  I  gives  the  point  at  which 
it  may  be  struck  so  that  the  hand  will  receive  no  jar. 

241.  The  Principal  Radius  of  Gyration  Deter- 
mined Practically. — Mount  the  body  upon  an  axis  not 
passing  through  the  centre  of  gravity,  and  cause  it  to 
oscillate  ;  from  the  number  of  oscillations  performed  in  a 
given  time,  say  an  hour,  the  time  of  one  oscillation  is 
known.  Then  to  find  //,  which  is  the  distance  from  the 
axis  to  the  centre  of  gravity,  attach  a  spring  balance  to  the 
lower  end,  and  bring  the  centre  of  gravity  to  a  horizontal 
plane  through  the  axis,  which  position  will  be  indicated  by 
the  maximum  reading  of  the  balance.  Knowing  the  maxi- 
mum reading,  E,  of  the  balance,  the  weight,  W,  of  the 
body,  and  the  distance,  «,  from  the  axis  of  suspension  to 


468  THE  BALLISTIC  PENDULUM. 

the  point  of  attachment,  we  have  from  the  principle  of 
moments,  Ra  =  Wh,  from  which  h  is  found.  Substitut- 
ing in  (3)  of  Art.  237,  this  value  of  h,  and  for  T  the  time 
of  an  oscillation,  Jcx  becomes  known. 

242.  The  Ballistic  Pendulum. — An  interesting  ap- 
plication of  the  principles  of  the  compound  pendulum  is 
the  old  way  of  determining  the  velocity  of  a  bullet  or  can- 
non-ball. It  is  a  matter  of  considerable  importance  in  the 
Theory  of  Gunnery  to  determine  the  velocity  of  a  bullet  as 
it  issues  from  the  mouth  of  a  gun.  It  was  to  determine 
this  initial  velocity  that  Mr.  Robins  about  1743  invented 
the  Ballistic  Pendulum.  This  consists  of  a  large  thick 
heavy  mass  of  wood,  suspended  from  a  horizontal  axis  in 
the  shape  of  a  knife-edge,  after  the  manner  of  a  compound 
pendulum.  The  gun  is  so  placed  that  a  ball  projected 
from  it  horizontally  strikes  this  pendulum  at  rest  at  a  cer- 
tain point,  and  gives  it  a  certain  angular  velocity  about  its 
axis.  The  velocity  of  the  ball  is  itself  too  great  to  be 
measured  directly,  but  the  angular  velocity  communicated 
to  the  pendulum  may  be  made  as  small  as  we  please  by 
increasing  its  bulk.  The  arc  of  oscillation  being  meas- 
ured, the  velocity  of  the  bullet  can  be  found  by  calcu- 
lation. 

The  time,  which  the  bullet  takes  to  penetrate,  is  so  short 
that  we  may  suppose  it  completed  before  the  pendulum  has 
sensibly  moved  from  its  initial  position. 

Let  M  be  the  mass  of  the  pendulum  and  ball;  m 
that  of  the  ball ;  v  the  velocity  of  the  ball  at  the  instant  of 
impact ;  h  the  distance  of  the  centre  of  gravity  of  the  pen- 
dulum and  ball  from  the  axis  of  suspension;  a  the  distance 
of  the  point  of  impact  from  the  axis  of  suspension  ;  w  the 
angular  velocity  due  to  the  blow  of  the  ball,  and  h  the 
radius  of  gyration  of  the  pendulum  and  ball.  Then  since 
the  initial  velocity  of  the. bullet  is  v,  its  impulse  is  measurec 
by  mv,  and  therefore  from  (3)  of  Art.  236  we  have  for  the 


THE  BALLISTIC  PENDUWUM,  4G9 

initial  angular  velocity  generated  in  the  pendulum  by  this 
impulse, 

mva 

Mk*' 


o,=:m:  (1) 


and   from    (1)   of  Art.  237   we   have   for   the  subsequent 
motion 

&0  9h      •       n  ( 

Integrating,  and  observing  that,  if  a  be  the  angle  through 
which  the  pendulum  moves,  we  have  — -  =  w  when  0  =  0, 

ttt 

rift 

and  —  =  0  when  6  =  a,  (2)  becomes 
tit 

a»  =  ^(l-cos«).  (3) 

Eliminating  o)  between  (1)  and  (3)  we  have 

2Mk    /-=-    .    « 

v  = Vgh  sin  ~,  (4) 

ma       J  2  v  ' 

from  which  v  becomes  known,  since  all  the  quantities  in 
the  second  member  may  be  observed,  or  are  known. 

We  may  determine  «  as  follows :  At  a  point  in  the  pen- 
dulum at  a  distance  h  from  the  axis  of  suspension,  attach 
the  end  of  a  tape,  and  let  the  rest  of  the  tape  be  wound 
tightly  round  a  reel ;  as  the  pendulum  ascends,  let  a  length 
c  be  unwound  from  the  reel ;  then  c  is  the  chord  of  the 

cc 

angle  a  to  the  radius  h,  so  that  c  =  2h  sin  «,  which  in  (4) 

gives 

Mice      lq 

V  =  \      \-  (5) 

ma    y  h  v  ' 

The  values  of  h  and  h  may  be  determined  as  in  Art.  241. 
If  the  mouth  of  the  gun  is  placed  near  to  the  pendulum, 


470  ROTATION  OF  A   HEAVY  BODY. 

the  value  of  v,  given  by  (5j,  must  be  nearly  the  velocity  of 
projection. 

The  velocity  may  also  be  determined  in  the  following 
manner :  Let  the  gun  be  attached  to  a  heavy  pendulum ; 
when  the  gun  is  discharged  the  recoil  causes  the  pendulum 
to  turn  round  its  axis  and  to  oscillate  through  an  arc 
w  hich  can  be  measured  ;  and  the  velocity  of  the  bullet  can 
be  deduced  from  the  magnitude  of  this  arc.  (See  Price's 
Anal.  Mech's,  Vol.  II,  p.  231.) 

Before  the  invention  of  tlie  ballistic  pendulum  by  Mr.  Robins  in 
1743,  but  little  progress  bad  been  made  in  the  true  theory  of  military 
projectiles.  Robins'  New  Principles  of  Gunnery  was  soon  translated 
into  several  languages,  and  Euler  added  to  his  translation  of  it  into 
German  an  extensive  commentary  ;  the  work  of  Euler's  being  again 
translated  into  English  in  1784.  The  experiments  of  Robins  were 
all  conducted  with  musket  balls  of  about  an  ounce  weight,  but  they 
were  afterwards  continued  during  several  years  by  Dr.  Hutton,  who 
used  cannon-balls  of  from  one  to  nearly  three  pounds  in  weight. 
Hutton  used  to  suspend  his  cannon  as  a  pendulum,  and  measure  the 
angle  through  which  it  was  raised  by  the  discharge.  His  experi- 
ments are  still  regarded  as  some  of  the  most  trustworthy  on  smooth 
bore  guns.  See  Routh's  Rigid  Dynamics,  p.  94,  also  Encyclopaedia 
Britannica,  Art.  Gunnery. 

243.  Motion  of  a  Heavy  Body  about  a  Horizon- 
tal Axle  through  its  Centre.— Let  the  body  be  a  sphere 
whose  radius  is  R,  and  weight  W,  and  let  a  weight  P  be 
attached  to  a  cord  wound  round  the  circumference  of  a 
wheel  on  the  same  axle,  the  radius  of  the  wheel  being  r ; 
required  the  distance  passed  over  by  P  in  t  seconds. 

From  (4)  of  Art.  236  we  have 

fflO  Prg 


W  ~~  Wk?  +  Pr* 

Multiplying  by  dt  and  integrating  twice,  we  have 

Prgfi  „. 

aOW  +  iV)'  lJ 


EXAMPLES. 


471 


the  constants  being  zero  in  both  integrations,  since  the  body 
starts  from  rest  when  t  =  0.    The  space  will  be  rd. 


EXAMPLES. 

1.  Let  the  body  be  a  sphere  whose  radius  is  3  ft.  and 
weight  500  lbs.;  let  P  be  50  lbs.,  and  the  radius  of  the 
wheel  6  ins.;  required  the  time  in  which  the  weight  P  will 
(Take#  =  32.) 

Aiis.  21  seconds. 


descend  through  50  ft 


2.  Let  the  body  be  a  sphere  whose  radius  is  14  ins.  and 
weight  800  lbs.;  let  it  be  moved  by  a  weight  of  200  lbs. 
attached  to  a  cord  wound  round  a  wheel  the  radius  of 
which  is  one  foot ;  find  the  number  of  revolutions  of  the 
sphere  in  eight  seconds.     (Take  g  =  32.)         Ans.  51.3. 


244.  Motion  of  a  Wheel  and 
Axle  when  a  Given  Weight  I* 
Raises    a   Given   Weight    W. — Let 

the  weights  P  and  W  be  attached  to 
cords  wound  round  the  wheel  and  axle, 
respectively,  (Fig.  97)  ;  let  R  and  r  be 
be  the  radii  of  the  wheel  and  axle,  and 
to  and  to'  their  weights;  required  the 
angular  distance  passed  over  in  t 
seconds. 
From  (4)  of  Art.  236,  we  have 

d26  PR  —  Wr 


Fig.97 


dfi  ~~  PR2  +  Wf*  +  \wR*  +  $w'r* 
(PR  _  Wr)  t* 


ffi 


PR?  +  ¥Pr«  +  \w&  +  iw'f* 


fr. 


(1) 

(2) 


EXAM  PLE, 


Let  the  weight  P  =  30  lbs.,  W  =  80  lbs.,  to  =  8  lbs., 
and  w  =  4  lbs.;  and  let  R  and  r  be  10  ins.  and  4  ins.; 


472  MOTION  ABOUT  A    VERTICAL   AXIS. 

required  (1)  the  space  passed  over  by  P  in  12  seconds  if  ii 
starts  from  rest,  and  (2)  the  tensions  Tand  T'  of  the  cords, 
supporting  P  and  W,     (Take  g  =  32.) 

Ans.   (1)  97.79  ft.;  (2)  T=  31.28  lbs.;  T  =  78.64  lbs. 

245.  Motion  of  a  Rigid  Body 
about  a  Vertical  Axis. — Let  AB  ^^^ 

be  a  vertical  axis  about  which  the       ' r^H 

body  C,  on  the  horizontal  arm  ED,     A 

revolves,  under  the  action  of  a  con-        % 

stant  horizontal  force  F,  applied  at  Fig.98 

the  extremity   E,  perpendicular   to 

ED.     Let  M  be  the  mass  of  the  body  whose  centre  is  C, 

and  r  and  h  the  distances  ED  and  CD,  respectively.     Then 

from  (4)  of  Art  236,  we  have 

m  Ft 


Multiplying  by  dt  and  integrating  twice,  observing  that 
the  constants  of  both  integrations  are  zero,  we  have 

which  is  the  angular  space  passed  over  in  t  seconds. 


EXAM  PLE. 

Let  the  body  be  a  sphere  whose  radius  is  2  ft.,  whose 
weight  is  600  lbs.,  and  the  distance  of  whose  centre  from 
the  axis  is  8  ft.,  and  let  F  be  a  force  of  40  lbs.  acting  at  the 
end  of  an  arm  10  ft.  long;  find  (1)  the  number  of  revolu- 
tions which  the  body  will  make  about  the  axis  in  10 
minutes,  and  (2)  the  time  of  one  revolution.  (Take 
g  =  32.)  Ans.   (1)  9316.3  ;  (2)  6.2  sees. 


/ 


Fig.99 


BODY  ROLLING   DOWN  AN  INCLINED   PLANE.        473 

246.  Body  Rolling  down  an   Inclined  Plane.— A 

homogeneous  sphere  rolls  directly  doivn  a  rough  inclined 
plane  under  the  action  of  gravity.     Find  the  motion. 

Let  Fig.  99  represent  a  section 
of  the  sphere  and  plane  made  by  a 
vertical  plane  passing  through  C, 
tha  centre  of  the  sphere.  Let  a  be 
the  inclination  of  the  plane  to  the 
horizon,  a  the  radius  of  the  sphere, 
0  the  point  of  the  plane  which 
was  initially  touched  by  the  sphere 

at  the  point  A,  P  the  point  of  contact  at  the  time  t, 
ACP  =  0,  which  is  the  angle  turned  through  by  the 
sphere,  m  =  the  mass  of  the  sphere,  F  =  the  frictioD 
acting  upwards,  R  =  the  pressure  of  the  sphere  on  the 
plane.  Then  it  is  convenient  to  choose  0  for  origin  and 
OB  for  the  axis  of  x ;  hence  OP  =  x. 

The  forces  which  act  on  the  sphere  are  (1)  the  reaction, 
R,  perpendicular  to  OB  at  P,  (2)  the  friction,  F,  acting  at 
P  along  PO,  and  (3)  its  weight,  mg,  acting  vertically  at  0 
the  centre.  Now  C  evidently  moves  along  a  straight  line 
parallel  to  the  plane ;  so  that  for  its  motion  of  translation 
we  have,  by  resolving  along  the  plane, 


m 


(Px 


mg  sin  a  —  F, 


(i) 


The  sphere  evidently  rotates  about  its  point  of  contact 
with  the  plane ;  but  it  may  be  considered  as  rotating  at 
any  instant  about  its  centre  C  as  fixed  ;  and  the  angular 
velocity  of  C  at  that  instant  in  reference  to  P  is  the  same 
as  that  of  P  m  reference  to  C.  From  (4)  of  Art.  236,  we 
have  for  the  motion  of  rotation 


mki' 


dt* 


=  Fa 


(2) 


474        BODY  ROLLING   DOWN  AN  INCLINED   PLANE. 

and  since  the  plane  is  perfectly  rough,  so  that  the  sphere 
does  not  slide,  we  have 

x  =  ad.  (3) 

Multiplying  (1)  by  a  and  adding  the  result  to  (2),  we  get 

ma7H*  +  m  l  If2  =  ma9  sm  a'  '  ■ 

Differentiating  (3)  twice  we  get  ^  =  a  -^,  which 
united  to  (4)  gives 

d2x  a2  ,„. 

^  =  «*-+*?  ^sma-  (5) 

Since  the  sphere  is  homogeneous,  kt2  =  \a2,  and  (5) 
becomes 

-^  =  \g  sin  a  (6) 

which  gives  the  acceleration  doivn  the  plane. 

If  the  sphere  had  been  sliding  down  a  smooth  plane,  the 
acceleration  would  have  been  g  sin  a  (Art.  144) ;  so  that 
two-sevenths  of  gravity  is  used  in  turning  the  sphere,  and 
five-sevenths  in  urging  the  sphere  down  the  plane. 

Integrating  (6)  twice,  and  supposing  the  sphere  to  start 
from  rest,  we  have 

x  =  -f-^g  -  sin  a  •  t2 

which  gives  the  space  passed  over  in  the  time  t. 
Resolving  perpendicular  to  the  plane,  we  have 

R  =  mg  cos  a. 

Cor. — If  the  rolling  body  were  a  circular  cylinder  with 
its  axis  horizontal,  then  Tc^  =  \a2,  and  (5)  becomes 

^  =  \s  sin  a ;  (7) 


IMPULSIVE  FORCE.  475 

so  that  one-third  of  gravity  is  used  in  turning  the  cylinder, 
and  two-thirds  in  urging  it  down  the  plane. 
From  (7)  we  have 

x  =  \g  sin  a  •  t2  (8) 

which  gives  the  space  passed  over  in  the  time  t  from  rest. 

247.  Motion  of  a  Falling  Body  under  the  Action 
of  an  Impulsive  Force  not  Directly  through  its 
Centre. — A  string  is  tvound  round  the  circumference  of  a 
reel,  and  the  free  end  is  attached  to  a  fixed  point.  The  reel 
is  then  lifted  up  and  let  fall  so  that  at  the  moment  when 
the  string  becomes  tight  it  is  vertical,  and  tangent  to  the  reel. 
The  whole  motion  being  supposed  to  take  place  in  one  plane, 
determine  the  effect  of  the  impulse. 

The  reel  at  first  will  fall  vertically  without  rotation. 
Let  v  be  the  velocity  of  the  centre  at  the  moment  when  the 
string  becomes  tight;  v' ,  (o  the  velocity  of  the  centre  and 
the  angular  velocity  just  after  the  impulse ;  T  the  impul- 
sive tension ;  m  the  mass  of  the  reel,  and  a  its  radius. 

Just  after  the  impact  the  part  of  the  reel  in  contact  with 
the  string  has  no  velocity,  and  at  this  instant  the  reel 
rotates  about  this  part;  but  it  may  be  considered  as 
rotating  about  its  axis  as -fixed,  and  the  angular  velocity  of 
its  axis,  at  this  instant,  in  reference  to  the  part  in  contact 
is  the  same  as  that  of  the  latter  in  reference  to  the  former. 
The  impulsive  tension  is 

T=m(v  —  v').  (1) 

Hence  from  (3)  of  Art.  236,  we  have  for  the  motion  of 
rotation 

mk^o)  =  m(v  —  v')a.  (2) 


476  IMPULSIVE  FORCE. 

Since  the  part  of  the  reel  in  contact  with  the  string  has 
no  velocity  at  the  instant  of  impact,  we  have 

v  =  a<o.  (3) 

Solving  (2)  and  (3)  we  have 

av 


co  = 


a?  +  kx* 


(4) 


a2 
If  the  reel  be  a  homogeneous  cylinder,  kt2  =  — ,  and  we 


have  from  (3)  and  (4) 


"  =  *!;>     "'  =  &,  (5) 

and  from  (1)  we  have  for  the  impulsive  tension, 
T  =  \mv. 

Cor. — To  find  the  subsequent  motion.  The  centre  of  the 
reel  begins  to  descend  vertically ;  and  as  there  is  no  hori- 
zontal force  on  it,  it  will  continue  to  descend  in  a  vertical 
straight  line,  and  throughout  all  its  subsequent  motion  the 
string  will  be  vertical.  The  motion  may  therefore  be 
easily  investigated,  as  in  Art.  246,  since  it  is  similar  to  the 
case  of  a  body  rolling  down  an  inclined  plane  which  is 
vertical,  the  tension  of  the  string  taking  the  place  of  the 

friction  along   the    plane.     Hence   putting    a  =  -,   and 

letting  the  friction  F  =  the  finite  tension  of  the  string,  we 
have,  from  (1)  and  (7)  of  Art.  246, 

F  =  \mg,     and    -^  =  \g  ; 
that  is,  the  finite  tension  of  the  string  is  one-third  of  the 


EXAMPLES.  477 

freight,  and  the  reel  descends  with  a  uniform  acceleration 
offc. 

Since  the  initial  velocity  of  the  reel  from  (5)  is  f  v,  we 
have,  for  the  space  descended  in  the  time  t  after  the  impact, 
from  (8)  of  Art.  246, 

x  =  fi't+  \gt2.     (See  Routh's  Rigid  Dynamics,  p.  131.) 

KX  A  m  p  le  s. 

1.  A  thin  rod  of  steel  10  ft.  long,  oscillates  about  an  axis 
passing  through  one  end  of  it ;  find  (1)  the  time  of  an 
oscillation,  and  (2)  the  number  of  oscillations  it  makes  in  a 
day.  Ans.   (1)  1.434  sec;  (2)  60254. 

~  2.  A  pendulum  oscillates  about  an  axis  passing  through 
its  end ;  it  consists  of  a  steel  rod  60  ins.  long,  with  a  rect- 
angular sectioii  \  by  J  of  an  inch ;  on  this  rod  is  a  steel 
cylinder  2  in.  in  diameter  and  4  in.  long;  when  the  ends  of 
the  rod  and  cylinder  are  set  square,  find  the  time  of  an 
oscillation.  Ans.  1.174  sees. 

3.  Determine  the  radius  of  gyration  with  reference  to 
the  axis  of  suspension  of  a  body  that  makes  73  oscillations 
in  2  minutes,  the  distance  of  the  centre  of  gravity  from  the 
axis  being  3  ft.  2  in.  Ans.  5.267  ft. 

4.  Determine  the  distance  between  the  centres  of  suspen- 
sion and  oscillation  of  a  body  that  oscillates  in  2J  sec. 

Ans.  20.264  ft. 

5.  A  thin  circular  plate  oscillates  about  an  axis  passing 
through  the  circumference ;  find  the  length  of  the  simple 
equivalent  pendulum,  (1)  when  the  axis  touches  the  circle 
and  is  in  its  plane,  and  (2)  when  it  is  at  right  angles  to 
the  plane  of  the  circle.  Ans.  (1)  Ja  ;  (2)  \a. 

6.  A  cube  oscillates  about  one  of  its  edges;  find  the 
length  of  the  simple  equivalent  pendulum,  the  edge  being 

=  2«-  Ans.  $a  a/2. 


478  EXAMPLES. 

7.  A  prism,  whose  cross  section  is  a  square,  each  side 
being  =  2a,  and  whose  length  is  Z,  oscillates  about  one  of 
its  upper  edges ;  find  the  length  of  the  simple  equivalent 
pendulum,  Ans.  f  V^a2  +  K 

8.  An  elliptic  lamina  is  such  that  when  it  swings  about 
one  latus  rectum  as  a  horizontal  axis,  the  other  latus 
rectum  passes  through  the  centre  of  oscillation ;  prove 
that  the  eccentricity  is  -J. 

9.  The  density  of  a  rod  varies  as  the  distance  from  one 
end ;  find  the  axis  perpendicular  to  it  about  which  the 
time  of  oscillation  is  a  minimum,  I  being  the  length  of  the 
rod. 

Ans.  The  distance  of  the  axis  from  the  centre  of  gravity 

10.  Find  the  axis  about  which  an  elliptic  lamina  must 
oscillate  that  the  time  of  oscillation  may  be  a  minimum. 

Ans.  The  axis  must  be  parallel  to  the  major  axis,  and 
bisect  the  semi-minor  axis. 

11.  Find  the  centre  of  percussion  of  a  cube  which  rotates 
about  an  axis  parallel  to  the  four  parallel  edges  of  the  cube, 
and  equidistant  from  the  two  nearer,  as  well  as  from  the 
two  farther  edges.  Let  %a  be  a  side  of  the  cube,  and  let  c 
be  the  distance  of  the  rotation-axis  from  its  centre  of 
gravity. 

2a2 
Ans.  I  =  c  +  ■=-,  where  I  is  the  distance  from  the  rota- 
oc 

tion-axis  to  the  centre  of  percussion. 

12.  Find  the  centre  of  percussion  of  a  sphere  which 
rotates  about  an  axis  tangent  to  its  surface. 

Ans.  I  =  \a. 

13.  Let  the  body  in  Art.  243,  be  a  sphere  whose  radius  is 
17  ins.  and  weight  1200  lbs.;  let  it  be  moved  by  a  weight 
of  250  lbs.  attached  to  a  cord  wound  round  a  wheel  whose 


EXAMPLES.  479 

radius  is  15  ins. ;   find  the  number  of   revolutions  of  the 
sphere  in  10  seconds,     (g  =  32.)  Ans.  58.77. 

14.  Let  the  body  in  Art.  243  be  a  sphere  of  radius  8  ins. 
and  weight  500  lbs.;  let  it  be  moved  by  a  weight  of  100  lbs. 
attached  to  a  cord  wound  round  a  wheel  whose  radius  is 
6  in.;  find  the  number  of  revolutions  of  the  sphere  in 
5  seconds,     (g  =  32f )  Ans.  28.09. 

15.  In  Art.  244,  let  the  weight  P  =  40  lbs.,  W  =  100 
lbs.,  to  =  12  lbs.,  and  to'  =  6  lbs.;  and  let  R  and  r  be 
12  ins.  and  7  ins. ;  required  (1)  the  space  passed  over  by  P 
in  16  sees,  if  it  starts  from  rest,  and  (2)  the  tensions  T  and 
T'  of  the  cords  supporting  P  and  W.     (g  =  32). 

Ans.   (1)  926.5;  (2)  T  =  49.04  lbs.,  T'  =  86.81  lbs. 

16.  In  Art.  244,  let  the  weight  P  =  25  lbs.,  W  =  60 
lbs.,  to  =  6  lbs.,  and  to'  =  2  lbs. ;  and  let  R  and  r  be 
8  in.  and  3  in.;  required  (1)  the  space  passed  over  by  P  in 
10  sees,  if  it  starts  from  rest,  and  (2)  the  tensions  T  and 
T'  of  the  cords  supporting  P  and  W.     (g  ==  32£.) 

Ans.   (1)  109.92  ft.;  (2)  T=  23.29  lbs.;  T'=  61.54=  lbs. 

17.  In  Art.  245,  let  the  body  be  a  sphere  whose  radius 
is  3  ft.,  whose  weight  is  800  lbs.,  and  the  distance  of  whose 
centre  from  the  axis  is  9  ft. ;  and  let  i^bea  force  of  60  lbs. 
acting  at  the  end  of  an  arm  12  ft.  long;  find  (1)  the  num- 
ber of  revolutions  which  the  body  will  make  about  the 
axis  in  12  min.,  and  (2)  the  time  of  one  revolution. 
(g  =  32.)  Ans.   (1)  14043.6  ;  (2)  6.07  sees. 

18.  In  Ex.  17,  let  the  radius  =  one  foot,  the  weight  = 
100  lbs.,  the  distance  of  centre  from  axis  =  5  ft.,  and 
F  =  25  lbs.  acting  at  end  of  arm  8  ft.  long;  find  (1)  the 
number  of  revolutions  which  the  body  will  make  about  the 
axis  in  5  min.,  and  (2)  the  time  of  one  revolution. 
(g  —  32f )  Ans.   (1)  18139.09  ;  (2)  2.23  sees. 

19.  If  the  body  in  Art.  247  be  a  homogeneous  sphere, 
the  string  being  round  the  circumference  of  a  great  circle, 


480  EXAMPLES. 

find  (1)  the  angular  velocity  just  after  the  impulse,  and  (2) 
the  impulsive  tension.  .        ov      ,  .  . 

20.  A  bar,  I  feet  long,  falls  vertically,  retaining  its  hori- 
zontal position  till  it  strikes  a  fixed  obstacle  at  one-quarter 
the  length  of  the  bar  from  the  centre  ;  find  (1)  the  angu- 
lar velocity  of  the  bar,  (2)  the  linear  velocity  of  its  centre 
just  after  the  impulse,  and  (3)  the  impulsive  force,  the 
velocity  at  the  instant  of  the  impulse  being  v. 

Ans.  (l)HJ';  (3)*>;  (3)  4pnv. 

21.  A  bar,  40  ft.  long,  falls  through  a  vertical  height  of 
50  ft.,  retaining  its  horizontal  position  till  one  end  strikes 
a  fixed  obstacle  60  ft.  above  the  ground  ;  find  (1)  its  angu- 
lar velocity,  (2)  the  linear  velocity  of  its  centre  just  after 
the  impulse;  (3)  the  number  of  revolutions  it  will  make 
before  reaching  the  ground,  (4)  the  whole  time  of  falling 
to  the  ground,  and  (5)  its  linear  velocity  on  reaching  the 
ground. 

A?is.  (1)  2.12;  (2)  42.43;  (3)  0.345;  (4)  2.79;  (5) 
75.10. 


CHAPTER    VIII. 

MOTION    OF    A    SYSTEM    OF    RIGID    BODIES  IN   SPACE 

248.  The   Equations   of  Motion  of  a   System  of 
Rigid  Bodies  obtained  by  D'Alembert's  Principle. — 

Let  (x,  y,  z)  be  the  position  of  the  particle  m  at  the  time  t 
referred  to  any  set  of  rectangular  axes  fixed  in  space,  and 
X,  Y,  Z,  the  axial  components  of  the  impressed  accelera- 

fPcc    fl21l    fl2Z 

ting  forces  acting  on  the  same  particle.     Then  -r-r,  -~9  ~—n, 

°  °  at2    at2    dt2 

are  the  axial  components  of  the  accelerations  of  the  parti- 
cle ;  and  by  D'Alembert's  Principle  (Art.  235)  the  forces, 


m 


(*-£)■  -('-39-  •(*-» 


acting  on  m  together  with  similar  forces  acting  on  every 
particle  of  the  system,  are  in  equilibrium.  Hence  by  the 
principles  of  Statics  (Art.  65)  we  have  the  following  six 
equations  of  motiou  : 

Zm  (*X-  mt)  -  2m  (/g  -  zg)  =  0,J         (2) 

a.(.F_,ri)-2»(.gr_,g)  =  a 

21 


482  TRANSLATION  AND   ROTATION. 

By  means  of  these  six  equations  the  motion  of  a  rigid 
body  acted  on  by  any  finite  forces,  may  be  determined. 
They  lead  immediately  to  two  important  propositions,  one 
of  which  enables  us  to  calculate  the  motion  of  translation 
of  the  body  in  space ;  and  the  other  the  motion  of  rotation. 

249.  Independence  of  the  Motion  of  Translation 
of  the  Centre  of  Gravity,  and  of  Rotation  about  an 
Axis  Passing  through  it. — Let  (£,  y,  z)  be  the  position 
of  the  centre  of  gravity  of  the  body  at  the  time  t,  referred 
to  fixed  axes,  (x,  y,  z)  the  position  of  the  particle  m  referred 
to  the  same  axes,  (x',  y\  z')  the  position  of  m  referred  to  a 
system  of  axes  passing  through  the  centre  of  gravity  and 
parallel  to  the  fixed  axes,  and  M  the  whole  mass.    Then 

1.  x  =  x  +  x\    y  =  y  +  y\    z  =  z  +  /.  (1) 

Since  the  origin  of  the  movable  system  is  at  the  centre  of 
gravity,  we  have  (Art.  59) 

Zmx'  =  Zmy'  =  5W  =  0;  (2) 

Also     Zmx  =  Mx,     Zmy  =  My,  2mz  =  Mz\ 

d*x       „&'x     _    d2y        ^cPy     _    cPz       „&* 

Substituting  these  values  in  (1)  of  Art.  248,  we  have 


jrf|  =  s.f»r;> 


motiqx  of  a  body.  483 

These  three  equations  do  not  contain  the  co-ordinates  of 
the  point  of  application  of  the  forces,  and  are  the  same  as 
those  which  would  be  obtained  for  the  motion  of  the 
centre  of  gravity  supposing  the  forces  all  applied  at  that 
poiut.     Hence 

The  motion  of  the  centre  of  gravity  of  a  system  acted  on 
by  any  forces  is  the  same  as  if  all  the  mass  were  collected  at 
the  centre  of  gravity  and  all  the  forces  were  applied  at  that 
point  parallel  to  their  former  directions. 

2.  Differentiating  (1)  twice  we  have 


d2x 
dP 

+ 

dp 

'    dP 

= 

*9, 

dP  "*" 

cPyf 
dP* 

ePz 

di* 

= 

drz 
dP  + 

dP 

• 

Substituting  these  values  in  the  first  of  equations  (2)  ol 
Art.  248,  we  have 

Zm[Q,  +  y')Z-(z  +  z')Y] 

^»[>VrtffirS}-*+*)gt3)]-. 

Performing  the  operations  indicated  we  get 


484  TRANSLATION  AND   ROTATION. 

Omitting  the  1st,  2d,  4th,  5th,  Gth,  and  8th  terms  which 
vanish  hy  reason  of  (2),  (3),  and  (4),  we  have 

similarly  from  the  other  two  equations  of  (2)  we  havef 

These  three  equations  do  not  contain  the  co-ordinates  of 
the  centre  of  gravity,  and  are  exactly  the  equations  we 
would  have  obtained  if  we  had  regarded  the  centre  of 
gravity  as  a  fixed  point,  and  taken  it  as  the  origin  of 
moments.     Hence 


TJie  motion  of  a  tody,  acted  on  by  any  forces,  about  its 
centre  of  gravity  is  the  same  as  if  the  centre  of  gravity  were 
fixed  and  the  same  forces  acted  on  the  body.  That  is,  from 
(4)  the  motion  of  translation  of  the  centre  of  gravity  of  the 
body  is  independent  of  its  rotation  ;  and  from  (5)  the  rota- 
tion of  the  body  is  independent  of  the  translation  of  its 
centre. 

These  two  important  propositions  are  called  respectively, 
the  principles  of  the  conservation  of  the  motions  of  transla- 
tion and  rotation. 

Sch. — By  the  first  principle  the  problem  of  finding  the 
motion  of  the  centre  of  gravity  of  a  system,  however  com- 
plex the  system  may  be,  is  reduced  to  the  problem  of 
finding  the  motion  of  a  single  particle.  By  the  second 
principle  the  problem  of  finding  the  angular  motion  of 
free  body  in  space  is  reduced  to  that  of  determining  th 
motion  of  that  body  about  a  fixed  point 


CONSERVATION    OF   CENTRE   OF   GRAVITY.  485 

Rem. — In  using  the  first  principle  it  should  be  noticed 
that  the  impressed  forces  are  to  be  applied  at  the  centre  of 
gravity  parallel  to  their  former  directions.  Thus,  if  a  rigid 
body  be  moving  under  the  influence  of  a  central  force,  the 
motion  of  the  centre  of  gravity  is  not  generally  the  same 
as  if  the  whole  mass  were  collected  at  the  centre  of  gravity 
and  it  were  then  acted  on  by  the  same  central  force.  What 
the  principle  asserts  is,  that,  if  the  attraction  of  the  central 
force  on  each  element  of  the  body  be  found,  the  motion  of 
the  centre  of  gravity  is  the  same  as  if  these  forces  were 
applied  at  the  centre  of  gravity  parallel  to  their  original 
directions. 

250.  The  Principle  of  the  Conservation  of  the 
Centre  of  Gravity. — Suppose  that  a  material  system  is 
acted  on  by  no  other  forces  than  the  mutual  attractions  of 
its  parts ;  then  the  impressed  accelerating  forces  are  zero, 
which  give 

SJ=  27=  ZZ  =  0; 

therefore  from  (4)  of  Art.  249,  we  get 

dfi  ~    '    dp        '    dp""' 


dx  dy  n     cfz 


ux  uy  o      uz  /t\ 

-j4  =  v0  cos  a,     -~  =  v0  cos  ft    ^  =  vQ  cos  y.     (1) 


where  vQ  is  the  velocity  of  the  centre  of  gravity  when 
t  =  0,  and  «,  ft  y,  are  the  angles  which  its  direction  makes 
with  the  axes.  Therefore,  calling  v  the  velocity  of  the 
centre  of  gravity  at  the  time  t,  we  have 

v  =  y——dt* —  =  *•>  (2) 

which  is  evidently  constant. 


486  CONSERVATION  OF  AREAS. 

If  v0  =  0,  the  centre  of  gravity  remains  at  rest. 
Integrating  (1)  we  get 

x  =  v0t  cos  a  -J-  a,    y  =  v0t  cos  j3  +  b, 

e  =  v0t  cosy  +  c; 

x  —  a      y  —  b       z  —-  c 


cos  cc        cos  j3  "    cos  y 


(3) 


(a,  b,  c)  being  the  place  of  the  centre  of  gravity  of  the 
system  when  t  =  0.  As  (3)  are  the  equations  of  a  straight 
line  it  follows  that  the  motion  of  the  centre  of  gravity  is 
rectilinear. 

Hence  when  a  material  system  is  in  motion  under  the 
action  of  forces,  none  of  lohich  are  external  to  the  system, 
then  the  centre  of  gravity  moves  uniformly  in  a  straight  line 
or  remains  at  rest. 

Eem. — Thus  the  motion  of  the  centre  of  gravity  of  a 
system  of  particles  is  not  altered  by  their  mutual  collision, 
whatever  degree  of  elasticity  they  may  have,  because  a 
reaction  always  exists  equal  and  opposite  to  the  action.  If 
an  explosion  occurs  in  a  moving  body,  whereby  it  is  broken 
into  pieces,  the  line  of  motion  and  the  velocity  of  the 
centre  of  gravity  of  the  body  are  not  changed  by  the 
explosion  ;  thus  the  motion  of  the  centre  of  gravity  of  the 
earth  is  unaltered  by  earthquakes  ;  volcanic  explosions  on 
the  moon  will  not  change  its  motion  in  space.  The  motion 
of  the  centre  of  gravity  of  the  solar  system  is  not  affected 
by  the  mutual  and  reciprocal  action  of  its  several  members ; 
it  is  changed  only  by  the  action  of  forces  external  to  the 
system. 

251.  The  Principle  of  the  Conservation  of  Areas.— 

If  x,  y  be  the  rectangular,  and  r9  6  the  polar  co-ordinates 
of  a  particle,  we  have 


CONSERVATION   OF  AREAS,  487 

dy  _     dx_       d/y\ 
Xdt       y dt  ~X  dt\x) 

be  r»  cos2  6  jt  (tan  6?)  =  r*~.  (1) 

Now  \r*&Q  is  the  elementary  area  described  round  the 
origin  in  the  time  dt  by  the  projection  of  the  radius  vector 
of  the  particle  on  the  plane  of  xy,  (Art.  182.)  If  twice 
this  polar  area  be  multiplied  by  the  mass  of  the  particle, 
it  is  called  the  area  conserved  by  the  particle  in  the  time  dt 
round  the  axis  of  z.    Hence 


H>§->%) 


is  called  the  area  conserved  by  the  system. 

Let  dAx,  dAy,  dAs  be  twice  the  areas  described  by  the 
projections  of  the  radius  vector  of  the  particle  m  on  the 
planes  of  yz,  zx,  xy,  respectively ;  then  from  (1)  we  have 

_     /  dy         dx\       _    dA9 

and  differentiating  we  get 

_     /   6N         d?x\       ^     d*Az  ,A% 

if  the  impressed  accelerating  forces  are  zero  the  first 
member  of  (2)  is  zero,  from  (5)  of  Art.  249 ;  therefore  the 
second  member  is  zero.    Hence 


488  CONSERVATION   OF    VIS    VIVA. 

similarly        Sw^-O,    2m^  =  0; 

and  therefore  by  integration 

1m  d4^  =  h,    2m  -^  =  &',    Sro  ^  =  //', 
c?^  at  at 

h9  h',  h"  being  constants. 

.•.    lmAx  =z  ht,    imA9  =  h%    ZmAz  =  h"t; 

the  limits  of  integration  being  such  that  the  areas  and  the 
time  begin  simultaneously.  Thus,  the  sum  of  the  products 
of  the  mass  of  every  particle,  and  the  projection  of  the  area 
described  by  its  radius  vector  on  each  co-ordinate  plane, 
varies  as  the  time.  This  theorem  is  called  the  principle  of 
the  conservation  of  areas.    That  is, 

When  a  material  system  is  in  motion  under  the  action 
offerees,  none  of  which  are  external  to  the  system,  then  the 
sum  of  the  products  of  the  mass  of  each  particle  by  the  pro- 
jection, on  any  plane,  of  the  area  described  by  the  radius 
vector  of  this  particle  measured  from  any  fixed  point,  varies 
as  the  time  of  motion. 

252.  Conservation  of  Vis  Viva  or  Energy.* — Let 

(x,  y,  z)  be  the  place  of  the  particle  m  at  the  time  t,  and 
let  X,  Y,  Z  be  the  axial  components  of  the  impressed 
accelerating  forces  acting  on  the  particle,  as  in  Art.  248. 
The  axial  components  of  the  effective  forces  acting  on  the 
same  particle  at  any  time  t  are 

d2x         d?y         d?z 
mdf»     mJ>    m&- 

If  the  effective  forces  on  all  the  particles  be  reversed, 

*  See  Art.  189. 


CONSERVATION    OF    VIS    VIVA,  489 

they  will  be  in  equilibrium  with  the  whole  group  of  im- 
pressed forces  (Art.  235).  Hence,  by  the  principle  of 
virtual  velocities  (Art.  104),  we  have 


,m 


H*-a*+(r-30*+(*-3H*m 


where  dx,  dy,  6z  are  any  small  arbitrary  displacements  of 
the  particle  m  parallel  to  the  axes,  consistent  with  the  con- 
nection of  the  parts  of  the  system  with  one  another  at  the 
time  /. 

Xow  the  spaces  actually  described  by  the  particle  m  dur- 
ing the  instant  after  the  time  t  parallel  to  the  axes  are 
consistent  with  the  connection  of  the  parts  of  the  system 
with  each  other,  and  hence  w7e  may  take  the  arbitrary  dis- 
placements,  dx,   dy,   dz,    to  be  respectively   equal  to    the 

actual  displacements,  j-  dt,  —  dt,  ~  dt,  of  the  particle.* 
Making  this  substitution,  (1)  becomes 

t<$x  dx      &y  dy      &z  dz\ 

W  dt  +  cW  dt  +  dfi  dt) 

—  {*%+*% +  z7> 

Integrating,  wTe  get 

Zmf  —  S»u'02  =  2Swi  /  (Xdx  +  Ydy  -f  Zdz),     (3) 

^here  v  and  i\  are  the  velocities  of  the  particle  m  at  the 
times  t  and  t0. 

The  first  member  of  (2)  is  twice  the  vis  viva  or  kinetic 
energy  of  the  system  acquired  in  its  motion  from  the  time 
t0  to  the   time   t,  under   the  action  of  the  given  forces. 

*  That  is,  although  bx  is  not  equal  to  dx,  yet  the  ratio  of  Sx  to  dx  is  equal  to  the 
ratio  of  6t  to  dt. 


490  CONSERVATION    OF   VIS    VIVA. 

The  second  member  expresses  twice   the   work  done  b^ 
these  forces  in  the  same  time  (Art.  189). 

If  the  second  member  of  (2)  be  an  exact  differential  of  a 
function  of  x,  y,  z,  so  that  it  equals  df  (x,  y,  z) ;  then  tak- 
ing the  definite  integral  between  the  limits  x,  y,  z  and  xQ, 
y0,  z0,  corresponding  to  t  and  tQ,  (2)  becomes 

2mv*  -  2m V  =  2f{x,  y,  z)  -  2f(x0,  y0,  z0).      (3) 

Now  the  second  member  of  (2)  is  an  exact  differential  so 
far  as  any  particle  m  is  acted  on  by  a  central  force  whose 
centre  is  fixed  at  (a,  b,  c),  and  which  is  a  functior.  of  the 
distance  r  between  the  centre  and  (x,  y,  z)  the  place  of  m. 
Thus,  let  P  be  the  central  force  5=  /  (r),  say :  then 

r*  =  (x  —  af  +  (y  -  b)z  +  {z  —  cf\ 
,*.    rdr  =  {x  —  a)  dx  +  (y  —  l)  dy  +  (z  —  c)  dz; 
.•.    m  (Xdx  +  J7Z?/  +  Zdz)  =  mf(r)  dr; 

which  is  an  exact  differential ;    substituting  this  in  the 
second  member  of  (2),  it 


=  2mfrf(r)dr, 

To 


where  the  limits  r  and  r0  correspond  to  t  and  t0. 

Also,  the  second  member  of  (2)  is  an  exact  differential, 
so  far  as  any  two  particles  of  the  system  are  attracted 
towards  or  repelled  from  each  other  by  a  force  which  varies 
as  the  mass  of  each,  and  is  a  function  of  the  distance 
between  them.  Let  m  and  m'  be  any  two  particles  ;  let 
(x,  y,  z)f  (x',  y\  z')  be  their  places  at  the  time  t ;  r  their 
distance  apart;  P  =  f(r),  the  mutual  action  of  the  unit 
mass  of  each  particle.     Then  the  whole  attractive  force  of 


CONSERVATION    OF    VIS    VIVA.  491 

m  on  m'  is  Pm,  and  the  whole  attractive  force  of  m'  on  m 
is  Pm' ;  and  we  have 

X=m P,     Y=mZ ^- P,    Z=m P; 

r  r  r 

X'=-mX-—P9     Y'=-m£z£  P,    Z'=-mZ——R 
r  r  r 

Also      r2  =  (x  —  x'f  +  {y  —  #')*  +  (*  —  *')2- 
Therefore  for  these  two  particles,  we  have 
m  (Xdx  +  Ycly  -f  Ztfz)  +  m'  (IW  +  F%'  +  ^W) 

=  ^-'  P  [(*  -  a?')  (<fe  -  dx')  +  (y-  y')  [dy  -  dy') 

+  (*  -  «')  (efe  -  d*)] 

=  mm'f(r)  dr; 

which  is  an  exact  differential.  The  same  reasoning  applied 
to  every  two  particles  in  the  system  must  lead  to  a  similar 
result ;  so  that  finally  the  second  member  of  (2) 


=  2mm'  /  f  (r)  dr, 


where  the  limits  r  and  r0  correspond  to  t  and  t0,  so  that 
the  integral  will  be  a  function  solely  of  the  initial  and  final 
co-ordinates  of  the  particles  of  the  system. 

Hence,  when  a  material  system  is  in  motion  under  the 
action  of  forces,  none  of  which  are  external  to  the  system, 
then  the  change  of  the  vis  viva  of  the  system,  in  passing 
from  one  position  to  another,  depends  only  on  the  ttvo  posi- 
tions of  the  system,  and  is  independent  of  the  path  described 
by  each  particle  of  the  system. 

This  theorem  is  called  the  principle  of  the  conservation  of 
vis  viva  or  energy. 


492  PRINCIPLE    OF   VIS    VIVA. 

Cor.  1. — If  a  system  be  under  the  action  of  no  external 
forces,  we  have  X  =  Y  =  Z  =  0,  and  hence  the  vis  viva 
of  the  system  is  constant. 

Cor.  2. — Let  gravity  be  the  only  force  acting  on  the 
system.  Let  the  axis  of  z  be  vertical  and  positive  down- 
wards, then  we  have  X  =  0,  Y  =  0,  Z  =  g.  Hence  (2) 
becomes 

Zmv2  —  I,mv02  =  22m  (z  —  z0). 

But  if  z  and  z0  are  the  distances  from  the  plane  of  xy  to  the 
centre  of  gravity  of  the  system  at  the  times  t  and.  t0,  and  >f 
M  is  the  mass  of  the  system,  we  have 

Ml  =  Sfras,     Mz0  =  ZmzQ; 
.-.    Zmv2  —  Y>mv2  =  2%  (i  —  z0).  (4) 

That  is,  the  increase  of  vis  viva  of  the  system  depends  only 
on  the  vertical  distance  over  which  the  centre  of  gravity 
passes  ;  and  therefore  the  vis  viva  is  the  same  ivhenever  the 
centre  of  gravity  passes  through  a  given  horizontal  plane. 

Rem. — The  principle  of  vis  viva  was  first  used  by  Huyghens  in 
his  determination  of  the  centre  of  oscillation  of  a  body  (Art.  237, 
Rem.). 

The  advantage  of  this  principle  is  that  it  gives  at  once  a  relation 
between  the  velocities  of  the  bodies  considered  and  the  co-ordinates 
which  determine  their  positions  in  space,  so  that  when,  from  the 
nature  of  the  problem,  the  position  of  all  the  bodies  may  be  made  to 
depend  on  one  variable,  the  equation  of  vis  viva  is  sufficient  to  deter 
mine  the  motion. 

Suppose  a  weight  mg  to  be  placed  at  any  height  h  above  the  sur- 
face of  the  earth.  As  it  falls  through  a  height  z,  the  force  of  gravity 
does  work  which  is  measured  by  mgz.  The  weight  has  acquired  a 
velocity  v,  and  therefore  its  vis  viva  is  hnv1  which  is  equal  to  mgz 
(Art.  217).  If  the  weight  falls  through  the  remainder  of  the  height 
h,  gravity  does  more  work  which  is  measured  by  mg  {h  —  z).  When 
the  weight  has  reached  the  ground,  it  has  fallen  as  far  as  the  circum 


COMPOSITION   OF  ROTATIONS.  493 

(Stances  of  the  case  permit,  and  gravity  lias  done  work  which  is  meas- 
ured by  mgh,  and  can  do  no  more  work  until  the  weight  has  been 
lifted  up  again.  Hence,  throughout  the  motion  when  the  weight  has 
descended  through  any  space  z,  its  vis  viva,  l-mc-  (=  mgz),  together 
with  the  work  that  can  be  done  daring  the  rest  of  the  descent, 
mg  (h  —  z),  is  constant  and  equal  to  mgh,  the  work  done  by  gravity 
during  the  whole  descent  h. 

If  we  complicate  the  motion  by  making  the  weight  work  some 
machine  daring  its  descent,  the  same  theorem  is  still  true.  The  vis 
viva  of  the  weight,  when  it  has  descended  any  space  2,  is  equal  to  the 
work  mgz  which  has  been  done  by  gravity  daring  this  descent,  dimin- 
ished by  the  work  done  on  the  machine.  Hence,  as  before,  the  vis 
viva  together  with  the  difference  between  the  work  done  by  gravity 
and  that  done  on  the  machine  during  the  remainder  of  the  descent  is 
constant  and  equal  to  the  excess,  of  the  work  done  by  gravity  over 
that  done  on  the  machine  during  the  whole  descent.  (See  Routh's 
Rigid  Dynamics,  p.  270.) 

253.  Composition  of  Rotations. — It  is  often  neces- 
sary to  compound  rotations  about  axes  which  meet  at  a 
point.  When  a  body  is  said  to  have  angular  velocities 
about  three  different  axes  at  the  same  time,  it  is  only  meant 
that  the  motion  may  be  determined  as  follows :  Divide  the 
whole  time  into  a  number  of  infinitesimal  intervals  each 
equal  to  dt.  During  each  of  these,  turn  the  body  round 
the  three  axes  successively,  through  angles  u^lt,  u2dt,  (ozdt. 
The  result  will  be  the  same  in  whatever  order  the  rotations 
take  place.  The  final  displacement  of  the  body  is  the 
diagonal  of  the  parallelopiped  described  on  these  three  lines 
as  sides,  and  is  therefore  independent  of  the  order  of  the 
rotations.  Since  then  the  three  successive  rotations  are 
quite  independent,  they  may  be  said  to  take  place  simul- 
taneously. 

Hence  we  infer  that  angular  velocities  and  angular  accel- 
erations may  be  compounded  and  resolved  by  the  same 
rules  and  in  the  same  way  as  if  they  were  linear.  Thus,  an 
angular  velocity  w  about  any  given  axis  may  be  resolved 
into  two,  cj  cos  cc  and  w  sin  «,  about  axes  at  right  angles  to 


494 


Mui'IOX  OF  A    RIGID   BODY. 


each  other  and  making  angles  a  and  -  —  «  with  the  given 

axis. 

Also,  if  a  body  have  angular  velocities  w15  o)2,  w3  about 
three  axes  at  right  angles,  they  are  together  equivalent  to 
a  single  angular  velocity  g>,  where  co  =  Vw12  +  w22-|-aj32, 
about  an  axis  inclined  to  the  given  axes  at  angles  whose 

cosines  are  respectively  --,  — -,  — -• 


254.  Motion  of  a  Rigid  Body  referred  to  Fixed 
Axes. — Let  us  suppose  that  one  point  in  the  body  is  fixed. 
Let  this  point  be  taken  as  the  origin  of  co-ordinates,  and 
Jet  the  axes  OX,  OY,  OZ  be  any  directions  fixed  in  space 
and  at  right  angles  to  one  another.  The  body  at  the  time 
t  is  turning  about  some  axis  of  instantaneous  rotation 
(Art.  240).  Let  its  angular  velocity  about  this  axis  be  w, 
and  let  this  be  resolved  into  the  angular  velocities  »„  w2, 
t>3  about  the  co-ordinate  axes.     It  is  required  to  find  the 

resolved  linear  velocities,  -77,-77,-70,  parallel  to  the  axes  of 

dt    dt    at 

co-ordinates,  of  a  particle  m  at  the  point  P,  (x,  y,  z),  in 

terms  of  the  angular  velocities  about  the  axes. 

These  angular  velocities  are  sup- 
posed positive  when  they  tend  the 
same  way  round  the  axes  that 
positive  couples  tend  in  Statics 
(Art.  65).  Thus  the  positive 
directions  of  a19  w2,  w3  are  re- 
spectively from  y  to  z  about  x, 
from  z  to  x  about  y,  and  from  x 
to  y  about  z ;  and  those  negative 
which  act  in  the  opposite  direc- 
tions. 

Let  us  determine  the  velocity 
of  F  parallel  to  the  axis  of  z.     Let  PN  be  the  ordinate  z, 


Fifl.100 


AXIS   OF  INSTANTANEOUS  ROTATION.  495 

and  draw  PM  perpendicular  to  the  axis  of  x.  The  velocity 
of  P  due  to  rotation  about  OX  is  o)tPM.  Resolving  this 
parallel  to  the  axes  of  y  and  z,  and  reckoning  those  linear 
velocities  positive  which  tend  from  the  origin,  and  vice 
versa,  we  have  the  velocity 

along        MN  =  —  o)tP3I  cos  NPM  =  -—&>,#; 
and  along       NP  =  o1PM  sin  NPM  =  (oty. 

Similarly  the  velocity  due  to  the  rotation  about  OY  par- 
allel to  OX  is  u2z,  and  parallel  to  OZ  is  —  g)2x.  And  that 
due  to  the  rotation  about  OZ  parallel  to  OX  is  —  o)3y,  and 
parallel  to  OY  is  o)3x. 

Adding  together  those  velocities  which  are  parallel  to 
the  same  axes,  we  have  for  the  velocities  of  P  parallel  to 
the  axes  of  x,  y,  and  z,  respectively, 

tier 

w  =  a>sz-<o3y, 

dy 

^  =  (03X  -  Wj*,  ) 

dz 


255.  Axis  of  Instantaneous  Rotation. — Every  par- 
ticle in  the  axis  of  instantaneous  rotation  is  at  rest  relative 
to  the  origin ;  hence,  for  these  particles  each  of  the  first 
members  of  (1)  in  Art.  254,  will  reduce  to  zero,  and  we 
have 

a2z  —  G)3y  =  0, 

0)^  —  0^2  =  0,}  (1) 

<M)ty  —  (i)2x  =  0, 


496  ANGULAR    VELOCITY. 

which  are  the  equations  of  the  axis  of  instantaneous  rota* 
lion,  the  third  equation  being  a  necessary  consequence  of 
the  first  two  ;  hence, 

that  is,  the  instantaneous  axis  is  a  straight  line  passing 
through  the  origin  which  is  at  rest  at  the  instant  con- 
sidered ;  and  the  whole  body  must,  for  the  instant,  rotate 
about  this  line. 

Cor. — Denote  by  «,  j3,  y  the  angles  which  this  axis 
makes  with  the  co-ordinate  axes  x,  y,  z,  respectively, 
then  (Anal.  Geom.,  Art.  175)  we  have 


cos  «  = 


cos  0  = 


cos  y  = 


W  + V  + w»8' 


^2 . 

"_8_ 


which  gives  the  position  of  the  instantaneous  axis  in  terms 
of  the  angular  velocities  about  the  co-ordinate  axes. 

256.  The  Angular  Velocity  of  the  Body  about  the 
Axis  of  Instantaneous  Rotation. — The  angular  veloc- 
ity of  the  body  about  this  axis  will  be  the  same  as  that  of 
any  single  particle  chosen  at  pleasure.  Let  the  particle  be 
taken  on  the  axis  of  x  ;  if  from  it  we  draw  a  perpendicular, 
p,  to  the  instantaneous  axis,  then  the  distance  of  the  par- 
ticle from  the  origin  being  x,  we  have 


EULER'S  EQUATIONS.  497 


p  =  x  sin  a  =  os  \/l  —  cos2  «  =  x\     —     2      •  ^s  . 


Since,  for  this  particle,  y  =  0,  2  =  0,  Ave  have  from  (1) 
of  Art.  254,  for  the  absolute  velocity, 


and  hence,  for  the  angular  velocity  v,  we  have 


v  =  -  =  Vw,2  +  w22  +  <V> 

which  is  the  angular  velocity  required. 

257.  Euler's  Equations. — To  determine  the  genera\ 
equations  of  motion  of  a  body  about  a  fixed  point 

Let  the  fixed  point  O-be  taken  as  origin  ;  let  {x,  y,  z)  be 
the  place  of  any  particle  m,  at  the  time  t,  referred  to  any 
rectangular  axes  fixed  in  space,  and  let  Oxt,  Oyly  Ozt  be 
the  principal  axes  of  the  body  (Art.  231).  Differentiating 
(1)  of  Art.  254  with  respect  to  t,  we  have 

d?x         da*  d(o*  .  x  t  N 

dPz=Z~dt~y~dt+<°2  ^tV  ""  "^  "  Ws  i****—1"**)' 

(Py         du»         do.  .  .  .  . 

^f  =  x-^  -  *-jg£  +  o)3  (o9g  -o>3y)  -  o>t  (uxy  -  a9x), 

cPz  dur  da*  ,  x  ,  x 

^  =  y-T^-^-^  +  Wl  K*  -  wl«)  -   W2  (W2*  ~  W3#> 

Denoting  by  Z,  if,  JV,  the  first  terms  respectively  of  (2), 

cPx         d*ii 
(Art.  248),  and  substituting  the  above  values  of  -^  and  -^| 

in  the  last  of  these  equations,  we  get 


408  EtJLER'S  EQUATIONS. 

— Zrnxy  (a* ^—u z^  +  Zm  (x2—y2)  (*))(*) 2  (  *' 

— Zmyzo) !  w  3  +  Zmxzco  2  w  3 

The  other  two  equations  may  be  treated  in  the  same  way. 

The  coefficients  in  this  equation  are  the  moments  and 
products  of  inertia  of  the  body  with  regard  to  axes  fixed  in 
space  (Art.  224),  and  are  therefore  variable  as  the  body 
moves  about.  Let  u)x,  o)y,  o)g  be  the  angular  velocities  about 
the  principal  axes.  Since  the  axes  fixed  in  space  are  per- 
fectly arbitrary,  let  them  be  so  chosen  that  the  principal 
axes  are  coinciding  with  them  at  the  moment  under  con- 
sideration.    Then  at  this  moment  we  have  (Art.  232), 

Zmxy  =  0,    l^myz  =  0,    Zmzx  =  0 ; 

,  ,.,       .     do).       du)x 
also  (*)t  =  <*)x,  <*>2  =  My,  oi3  =  d)z ;  and  likewise  ~~-  =  -~9 

etc.*     Hence,  denoting  by  A,  B,  C,  the  moments  of  inertia 
about  the  principal  axes  (Art.  231),  (1)  becomes 

in  which  all  the  coefficients  are  constants;   and  similarly 
for  the  other  two  equations. 

Hence,  uniting  them  in  order,  and  retaining  the  letters 
w1?  wg,  G>3,  since   they  are  equal  to  %,  (oy,  <*)z,  the  three 

— -~  =  — -,  for  the  chancres  in  the  two  angular  velocities,  w,  and  wj,  during  a 
dt        dt 

given  small  time  after  the  axis  of  x,  coincides  with  the  axis  of  SB,  will  differ  only  by 

a  quantity  which  depends  upon  the  angle  passed  through  hy  the  axis  of  x\  during 

that  given   small  time ;  the  difference  between  wl  and  wx  will  therefore  be  an 

infinitesimal  of  the  second  order  and  therefore  their  derivatives  will  be  equal.    (See 

Pratt's  Mech.,  p.  428.    For  further  demonstration  of  this  equality,  the  student  ifl 

referred  to  Routh's  Rigid  Dynamics,  pp.  138  aud  189.) 


EULER'S   EQUATIONS.  499 

equations  of  motion  of  t lie  body  referred  to  the  principal 
axes  at  the  fixed  point  are 

A^-(B-C)  <*,<*,  =L, 

O^f-(A-B)o)lo1,=2f, 

1  iese  are  called  Euler's  Equations. 

Sch. — If  the  bod}'  is  moving  so  there  is  no  point  in  it 
which  is  fixed  in  space,  the  motion  of  the  body  about  its 
centre  of  gravity  is  the  same  as  if  that  point  were  fixed. 

It  is  clear  that,  instead  of  referring  the  motion  of  the 
body  to  the  principal  axes  at  the  fixed  point,  as  Euler  has 
done,  we  may  use  any  axes  fixed  in  the  body.  But  these 
are  in  general  so  complicated  as  to  be  nearly  useless. 

258.  Motion  of  a  Body  about  a  Principal  Axis 
through  its  Centre  of  Gravity. — If  a  body  rotate  about 
one  of  its  principal  axes  passing  through  the  centre  of 
gravity,  this  axis  will  suffer  no  pressure  from  the  centrifu- 
gal force. 

Let  the  oody  rotate  about  the  axis  of  z;  then  if  &)  be  its 
angnlar  velocity,  the  centrifugal  force  of  any  particle  m 
will  be  (Art.  198,  Cor.  1) 


muPp  =  mo)2  \/x2  -\-  y2, 

which  gives  for  the  x-  and  ^/-components  m<*)2x  and  muPy ; 
and  the  moments  of  these  forces  with  respect  to  the  axes  of 
y  and  x  are  for  the  whole  body 

Hmufixz,     and     ^Lm^yz. 


500  AXIS   OF  PERMANENT  ROTATION. 

But  these  are  each  equal  to  zero  when  the  axis  of  lotation 
is  a  principal  axis  (Art.  232) ;  hence,  the  centrifugal  force 
will  have  no  tendency  to  incline  the  axis  of  z  towards  the 
plane  of  xy.  In  this  case  the  only  effect  of  the  forces  moPx 
and  muPy  on  the  axis  is  to  move  it  parallel  to  itself,  or  to 
translate  the  body  in  the  directions  of  x  and  y.  But  the 
sum  of  all  these  forces  is 

I*m(*)2x    and     'Lmi^y, 

each  of  which  is  equal  to  zero  when  the  axis  of  rotation 
passes  through  the  centre  of  gravity ;  hence  we  conclude 
that,  when  a  body  rotates  about  one  of  its  principal  axes 
passing  through  its  centre  of  gravity,  the  rotation  causes  no 
'pressure  upon  the  axis. 

If  the  body  rotates  about  this  axis  it  will  continue  to 
rotate  about  it  if  the  axis  be  removed.  On  this  account  a 
principal  axis  through  the  centre  of  gravity  is  called  an 
axis  of  permanent  rotation.* 

Sch. — If  the  body  be  free,  and  it  begins  to  rotate  about 
an  axis  very  near  to  a  principal  axis,  the  centrifugal  force 
will  cause  the  axis  of  rotation  to  change  continually,  inas- 
much as  the  foregoing  conditions  cannot  obtain,  and  this 
axis  of  rotation  will  either  continually  oscillate  about  the 
principal  axis,  always  remaining  very  near  to  it,  or  else  it 
will  remove  itself  indefinitely  from  the  principal  axis. 
Hence,  whenever  we  observe  a  free  body  rotating  about  an 
axis  during  any  time,  however  short,  we  may  infer  that  it 
has  continued  to  rotate  about  that  axis  from  the  beginning 
of  the  motion,  and  that  it  will  continue  to  rotate  about  it 
for  ever,  unless  checked  by  some  extraneous  obstacle.  (See 
Young's  Mechs.,  p.  230,  also  Venturoli,  pp.  135  and  160.) 


*  Pratt's  Mechs.,  p.  422.    Called  also  a  natural  axis  of  rotation,  see  Young's 
Meche.,  p.  230  ;  also  an  invariable  axis,  see  Price's  Mechs.,  Vol.  II,  p.  257. 


VELOCITY  ABOUT  A   PRINCIPAL  AXIS.  501 

259.  Velocity  about  a  Principal  Axis  when  there 
are  no  Accelerating  Forces. — In  this  case  L  =  M  = 
N  =  0  in  (2)  of  Art.  257  ;  also  A,  B}  G  are  constant  for 
the  same  body ;  and  if  we  put 

B-G  _  G-A  __         A-B 

—  -r*     d —  —  tr,      79 =a  ±19 


A  B  G 

(2)  of  Art.  257  becomes 

do1  =  F(*)2u3dt,    do>2  =  G(o3G)tdt, 

do)3  =  Huxb)2dt. 

Put  WjWgWg^  =  d(p,  and  we  have  (1) 

o)1do)1  =  Fd(p,    G)2dG)2  =z  Gdcp,    o)3dd)3  =  Hd<J>; 

and  integrating,  we  get 

»i«  =  2ify  +  «2,   w22  =  2G<p  +  #2,   «32  =  2H(f>  +  c2.   (2) 

where  a,  b,  c  are  the  initial  values  of  colf  gj2,  w3 ;   hence 
from  (1)  and  (2) 

dt  =  *  (3) 

V(2i^>  +  a2)  (2(70  +  62)  (2£fy  +  c2)  V 

Suppose  now  the  body  begins  to  turn  about  only  one  of 
the  principal  axes,  say  the  axis  of  x,  with  the  angular 
velocity  a,  then  b  =  0,  c  =  0,  and  (3)  becomes 


2VGH  (j>V%F<f>  +  a? 
Eeplacing  2ify  -f  a2  by  its  value  wx2,  and  e?0  by  its  value 

-St— j  we  have 
F    ' 


dt 


d(ot 


VGH    wi 


2  —  //2  » 


502  THE  INTEGRAL    OF  EULER'S  EQUATIONS. 

and  integrating,  we  get 

2a     "wj-l-fl' 

wx  +  a  v  J 

The  constant  C  mnst  be  determined  so  that  when  t  =  0, 
g)x  is  the  initial  velocity  a;  hence  e2a(7  =0  or  C'=  -a, 
which  makes  the  first  member  of  (4)  zero  for  every  value 
of  /.  Hence,  at  any  time  /,  we  must  aave  w1  =  a;  and 
therefore  from  (2)  0  =  0,  and  w2  =  w3  =  0.  Conse- 
quently the  impressed  velocity  about  one  of  the  principal 
axes  of  rotation  continues  perpetual  and  uniform,  as  before 
shown  (Art.  258). 

260.  The  Integral  of  Euler's  Equations. — A  body 
revolves  about  its  centre  of  gravity  acted  on  by  no  forces  but 
such  as  pass  through  that  point ;  to  integrate  the  equations 
of  motion. 

As  the  only  forces  acting  on  the  body  are  those  which 
pass  through  its  centre  of  gravity,  they  create  no  moment 
of  rotation  about  an  axis  passing  through  that  centre;  and 
therefore  (2)  of  Art.  257  become 

the  principal  axes  being  drawn  through  the  centre  of 
gravity. 


THE   INTEGRAL    OF  EULER'S    EQUATIONS.  503 

Multiply  these  equations  severally   (1)   by  o^,  ws,  <i>3  ; 
and  (2)  by  A(D19  Bu)2,  Cg)3,  and  add ;  then  we  have 


(3) 


(3) 


integrating,  we  have 

Au*  +  Bu2*  +  Cw32  =  7i2 . 

where  /i2  and  h2  are  the  constants  of  integration. 
Eliminating  w32  from  (3),  we  have 

A  (A  -  C)  ut*  +  B(B-C)  a>22  =  k*  -  Ch*; 

'•    "»*  =  B(B-  c^-W-AiA-  C)^*];  (4) 

and  W32  =  Q^—^  [£2  -  ^2  -  -4  U  -  5)  *>,*].     (5) 

Substituting  these  values  of  w2  and  w3  in  the  first  of  equa- 
tions (1),  we  have 

&».  ,  [(A-Q)(A-B)(         »-o»  \ 

~W  +  \_  BO  V}~A(A-C)J 

which  is  generally  an  elliptic  transcendent,  and  so  does  not 
admit  of  integration  in  finite  terms.  In  certain  particular 
cases  it  may  be  integrated,  which  will  give  the  value  of  g>j 
in  terms  of  t3  and  if  this  value  be  substituted  in  (4)  and  (5), 


504       APPLICATION   OF  THE   GENERAL   EQUATIONS. 

the  values  of  w2  and  w3  in  terms  of  t  will  be  known,  and 
thus,  in  these  cases,  the  problem  admits  of  complete  solu- 
tion. 

Cor. — Let  w^,  cjy,  (*)g  be  the  axial  components  of  the 
initial  angular  velocity  about  the  principal  axes  when 
t  =  0;  then  integrating  the  first  of  (2),  and  taking  the 
limits  corresponding  to  t  and  0,  we  have 

A(ot*  +  Bc>22  +  Cto32  =  At**  +  Bu>?  +  Ca?.      (7) 

Let  «,  (3,  y  be  the  direction-angles  of  the  instantaneous 
axis  at  the  time  t  relative  to  the  principal  axes ;  so  that,  if 
a)  is  the  instantaneous  angular  velocity,  and  2wr?  is  the 
moment  of  inertia  relative  to  that  axis,  we  have  (Art.  253), 
w  j  =  W  cos  a,  6) 2  =  to  cos  /3,  to  3  =  to  cos  y,  which  sub- 
stituted in  (7),  gives 

Aax*  +  Buf  +  Co)*  =  to2  (A  cos2  a  +  B  cos2 0  +  (7 cos2  y) 

=  6*£m#*  (Art.  232,  Cor.) 

=  2m#2 

=  the  vis  viva  of  the  body; 

from  which  it  appears  that  the  vis  viva  of  the  body  is  con- 
stant throughout  the  whole  motion. 

Eem. — An  application  of  the  general  equations  of  rotatory 
motion  (Art.  257),  which  is  of  great  interest  and  impor- 
tance, is  that  of  the  rotatory  phenomena  of  the  earth  under 
the  action  of  the  attracting  forces  of  the  sun  and  the  moon, 
the  rotation  being  considered  relative  to  the  centre  of 
gravity  and  an  axis  passing  through  it,  just  as  if  the  centre 
of  gravity  was  a  fixed  point  (Art.  249,  Sch.)  ;  and  the 
problem  treated  as  purely  a  mathematical  one.  Also,  in 
addition  to  the  sun  and  the  moon,  the  problem  may  be 


EXAMPLES.  505 

extended  so  as  to  include  the  action  of  all  the  other  bodies 
whose  influence  affects  the  motion  of  the  earth's  rotation. 
In  fact  the  investigation  of  the  motion  of  a  system  of  bodies 
in  space  might  be  continued  at  great  length  ;  but  such 
investigations  would  be  clearly  beyond  the  limits  proposed 
in  this  treatise.  The  student  who  desires  to  continue  this 
interesting  subject,  is  referred  to  more  extended  works.* 

EXAMPLES. 

1.  A  hollow  spherical  shell  is  filled  with  fluid,  and  rolls 
down  a  rough  inclined  plane;  determine  its  motion. 

Let  M  and  M'  be  the  masses  of  the  shell  and  fluid 
respectively,  h  and  h'  their  radii  of  gyration  respectively 
about  a  diameter,  and  a  and  a'  the  radii  of  the  exterior  and 
interior  surfaces  of  the  shell ;  then  using  the  same  nota- 
tion as  in  Art.  246,  we  have 

(if  +  M' )  ^  =  (M  +  M')g  sin  a  -  F.  (1) 

As  the  spherical  shell  rotates  in  its  descent  down  the  plane, 
the  fluid  has  only  motion  of  translation;  so  that  the  equa- 
tion of  rotation  is 

J«*g  =  Fa.  (2) 

Multiplying  (1)  by  a2  and  (2)  by  a,  and  adding,  we  have 

[{M  +  M')  a*  +  MT$\  H  =  (M  +  M')  a*g  sin  «.    (3) 

If  the  interior  were  solid,  and  rigidly  joined  to  the  shell, 
the  equation  of  motion  would  be 

*  See  Price's  Meet's,  Vol.  n,  Pratt's  Mech's,  Routh's  Rigid  Dynamics,  La  Plaoe'a 
Mecanique  Celeste,  etc. 


506 


EXAMPLES. 


&X 


[(M+M')  a*  +  Mk*  +  M'k'*]  ^  =  (M+M1)  a2g  sin  a.  (4) 


Integrating  (3)  and  (4)  twice,  and  denoting  by  *  and  s'  the 
spaces  through  which  the  centre  moves  during  the  time  t 
in  these  two  cases  respectively,  we  have 


(M  +  M')a2  +  MB 


(5) 


so  that  a  greater  space  is  described  by  the  sphere  which  has 
the  fluid  than  by  that  which  has  the  solid  in  its  interior. 

If  the  densities  of  the  solid  and  the  fluid  are  the  same, 
we  have  from  (5),  by  Art.  233,  Ex.  14, 


W 


7a5 


2a'5 


(Price's  Anal.  Mechs.,  Vol.  II,  p.  368). 


2.   A  homogeneous  sphere  rolls  down   within   a  rough, 
spherical  bowl;  it  is  required  to  determine  the  motion. 

Let  a  be  the  radius  of  the  sphere,  and  b  the  radius  of  the 
bowl ;  and  let  us  suppose  the  sphere  to  be  placed  in  the 
bowl  at  rest.  Let  OCQ  =  <p, 
QPA  =  0,"  BCO  =  cc,  u  = 
the  angular  velocity  of  the 
ball  about  an  axis  through  its 
centre  P,  k  =  the  correspond- 
ing radius  of  gyration  ;  031  = 
x,  MP  =  y  ;  m  =  the  mass  of 
the  ball.    Then 


=  —  R  sin  cf>  +  /"cos  <p~, 


a) 


m 


dt2  ~~ 


R  cos  0  -f-  F  sin  <f>  —  mg 


(2) 


EXAMPLES.  50) 

Also    x  =    (5  —  «)  sin  0 ;    y  =  J  —  (5  —  a)  cos  0. 

g  =  (»-«)-*f +  (*-)-*©'  (5) 

^co80+-Bin0=(*-a)^.  (6) 


^0 

^2 


«  (b-a)df2=  F—  m9  sin  ft  (7) 


.Now  to  determine  the  angular  velocity  of  the  ball,  we  must 
estimate  the  angle  described  by  a  fixed  line  in  it,  as  PA, 
from  a  line  fixed  in  direction,  as  PM,  and  the  ratio  of  the 
infinitesimal  increase  of  this  angle  to  that  of  the  time  will 
be  the  angular  velocity  of  the  ball. 


_  dMPA    _d<t>       dO 
"""    W  ""      dt      ~~  dt  +  dt 


Since  the  sphere  does  not  slide,  ad  =  b  (a  —  <p]  j 


a  —  b  d(f) 

d(*)  _  a  —  b  dfy9 
dt  "      a      dP ' 


<B) 


from  (3),  (7),  and  (8)  we  get 
df 


^"a^M  =  -^si°0;  (9) 


508  EXAMPLES. 

...    (J  -  a)  (^)2=  ~g-  (cos  <p  -  cos  a).  (10) 

Substituting  (9)  in  (7)  we  have 

F  =  \mg  sin  0.  (11) 

Substituting  (4),  (9),  (10),  (11)  in  (1)  we  have 

B  =  ^  (17  cos  ^  —  10  cos  «)  ; 

therefore  the  pressure  at  the  lowest  point 

=  !M(i7_i0cos«); 

and  the  pressure  of  the  ball  on  the  bowl  vanishes  when 

cos  (f>  =  \%  cos  a. 

Cor. — If  the  ball  rolls  over  a  small  arc  at  the  lowest  part 

of  the  bowl,  so  that  a  and  <p  are  always  small,  cos  «,  and 

a2  <p2 

cos  0  may  be  replaced  by  1  —  —  and  1  —  —  respectively ; 

and  from  (10)  we  have 

(&  _  08)4       l_7  {b  -  a) J 
...    0  =  «CoS[r(^J<; 

thus  the  ball  comes  to  rest  at  points  whose  angular  distance 
is  «  on  both  sides  of  0,  the  lowest  point  of  the  bowl ;  and 
the  periodic  time  is 


EX.  1  MFLES. 


509 


therefore  the  oscillations  are  performed  in  the  same  time  as 
those  of  a  simple  pendulum  whose  length  is  -}  (b  —  «), 
(Art.  194).     (Price's  Anal.  Mech's,  Vol.  II,  p.  3G9.) 

3.  A  homogeneous  sphere  has  an  angular  velocity  cj 
about  its  diameter,  and  gradually  contracts,  remaining 
constantly  homogeneous,  till  it  has  half  the  original 
diameter;  required  the  final  angular  velocity.     Ans.  4w. 

4.  If  the  earth  were  a  homogeneous  sphere,  at  what  point 
must  it  be  struck,  that  it  may  receive  its  present  velocity 
of  translation  and  of  rotation,  the  former  being  68000  miles 
per  hour  nearly  ?     Ans.  24  miles  nearly  from  the  centre. 

5.  A  homogeneous  sphere  rolls  down  a  rough  inclined 
plane;  the  inclined  plane  rests  on  a  smooth  horizontal 
plane,  along  which  it  slides  by  reason  of  the  pressure  of  the 
sphere;  required  the  motions  of  the  inclined  plane  and  of 
the  centre  of  the  sphere. 

Let  m  =  the  mass  of  the  sphere, 
M  =  the  mass  of  the  inclined 
plane,  a  =  the  radius  of  the  sphere, 
a  =  the  angle  of  the  inclined 
plane,  Q  its  apex ;  0  the  place  of  Q 
when  t  =  0;  O'  the  point  on  the 
plane  wrhich  was  in  contact  with 
the  point  A  of  the  sphere  when 
/  =  0,  at  wrhich  time  we  may  sup- 
pose all  to  be  at  rest ;  ACP  =  6,  the  angle  through  which 
the  sphere  has  revolved  in  the  time  L 

Let  O  be  the  origin,  and  let  the  horizontal  and  vertical 
lines  through  it  be  the  axes  of  x  and  y ;  OQ  =  x' ;  and  let 
(x,  y)  (h,  Tc)  be  the  places  of  the  centre  of  the  sphere  at  the 
times  ^  =  ^and/  =  0  respectively.  Then  the  equations 
of  motion  of  the  sphere  are 


Fig.  102 


m 


d?x 
dt2 


F  cos  a  —  R  sin  cc, 


510  EXAMPLES. 

m  -rjj-  =  F  sin  «  +  R  cos  «  —  mgs 

$ma*S = aF' 

and  the  equation  of  motion  of  the  plane  is 

d2x' 
M-Tp  =  —  Fcos  a  -j-  R  sin  «. 

From  the  geometry  we  have 

x  =  h  +  x'  —  ad  cos  a, 

y  =  h  —  ad  sin  a. 

From  these  equations  we  obtain 

,       mcos«    . 

x   = i>«0 

m  +  M 

-         5?ft  sm  «  cos  a  g& 

~~  7  (m  4-  J/)  —  5?/?  cos2  a "  2  ' 

5ilf  sin  cc  cos  «  otf2 

4«#      y  ^ .  «z 

7  (w  +  i¥)  —  5w  cos2  a     2  9 

—  h  _         5  (?/?  +  If)  sin2  cs  ^ 

^  ~~  7  (m  +  i¥)  —  5m  cos8  a  '  2 

which  give  the  values  of  #  and  y  in  terms  of  L 
Also  W3  obtain 

(m  -f-  J/)  (x  —  h)  sin  «  —  M  (y  —  h)  cos  a  =  0 ; 

which  is  the  equation  of  the  path  described  by  the  centre 
of  the  sphere  ;  and  therefore  this  path  is  a  straight  line. 

6.  A  heavy  solid  wheel  in   the  form  of  a  right  circulai 
cylinder,  is  composed  of  two  substances,  whose  volumes  are 


EXAMPLES,  5H 

equal,  and  whose  densities  are  p  and  p';  these  substances 
are  arranged  in  two  different  forms;  in  one  case,  that  whose 
density  is  p  occupies  the  central  part  of  the  wheel,  and  the 
other  is  placed  as  a  ring  round  it;  in  the  second  case,  the 
places  of  the  substances  are  interchanged  ;  t  and  t'  are  the 
times  in  which  the  wheels  roll  down  a  given  rough  inclined 
plane  from  rest;  show  that 

P  :  t'*  ::  5/>  +  7p'  :  op'  +  1P. 

7.  A  homogeneous  sphere  moves  down  a  rough  inclined 
plane,  whose  angle  of  inclination  «  to  the  horizon  is  greater 
than  that  of  the  angle  of  friction  ;  it  is  required  to  show  (1) 
that  the  sphere  will  roll  without  sliding  when  \i  is  equal  to 
or  greater  than  f  tan  «,  and  (2)  that  it  will  slide  and  roll 
when  \i  is  less  than  -|  tan  «,  where  \i  is  the  coefficient  of 
friction. 

8.  In  the  last  example  show  that  the  angular  velocity  of 
the  sphere  at  the  time  t  from  rest  =  — ^ 1. 

9.  If  the  body  moving  down  the  plane  is  a  circular 
cylinder  of  radius  =  a,  with  its  axis  horizontal,  show  that 
the  body  will  slide  and  roll,  or  roll  only,  according  as  «  is 
greater  or  not  greater  than  ta;1- !  3/u. 


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Child,  C.  T.     The  How  and  Why  of  Electricity i2mo,  1  00 

Christie,  W.  W.      Boiler-waters,  Scale,  Corrosion,    Foaming 

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Furnace  Draft.     (Science  Series) i6mo,  o  50 

— —  Water,  Its  Purification  and  Use  in  the  Industries ....  8vo, 

Church's  Laboratory  Guide.     Rewritten  by  Edward  Kinch.  ,8vo,  *2  50 

Clanperton,  G.     Practical  Papermaking 8vo,  2  50 

Clark,  A.  G.     Motor  Car  Engineering. 

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Clark,  D.  K.     Rules,  Tables  and  Data  for  Mechanical  Engineers 

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Fuel:  Its  Combustion  and  Economy i2mo,  1  50 

The  Mechanical  Engineer's  Pocketbook i6mo,  2  00 

Tramways :  Their  Construction  and  Working 8vo,  5  00 

Clark,  J.  M.     New  System  of  Laying  Out  Railway  Turnouts.. 

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i2mo,  *5  00 

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Cleemann,  T.  M.     The  Railroad  Engineer's  Practice i2mo,  *i  50 

Clerk,  D.,  and  Idell,  F.  E.     Theory  of  the  Gas  Engine.     (Science 

Series  No.  62.) i6mo,  o  50 

Clevenger,  S.  R.     Treatise  on  the  Method  of  Government  Sur- 
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Cochran,  J.     Traatise  on  Cement  Specifications. 8vo,  *i  oo 

Coffin,  J.  H.  C.     Navigation  and  Nautical  Astronomy i2mo,  *3  50 

Colburn,  Z.,  and  Thurston,  R.  H.     Steam  Boiler  Explosions. 

(Science  Series  No.  2.) i6mo,  o  50 

Cole,  R.  S.     Treatise  on  Photographic  Optics i2mo,  1  50 

Coles-Finch,  W.     Water,  Its  Origin  and  Use 8vo,  *5  00 

Collins,  J.  E.     Useful  Alloys  and  Memoranda  for  Goldsmiths, 

Jewelers i6mo,  o  50 

Constantine,  E.     Marine  Engineers,  Their    Qualifications    and 

Duties 8vo,  *2  00 

Coombs,  H.  A.     Gear  Teeth.     (Science  Series  No.  120).  .  .i6mo,  o  50 

Cooper,  W.  R.     Primary  Batteries 8vo,  *4  00 

"  The  Electrician  "  Primers 8vo,  *5  00 

Part  I *i  50 

Part  II *2  50 

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Corfield,  W.  H.  Dwelling  Houses.  (Science  Series  No.  50.)  i6mo,  0  50 

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Cornwall,  H.  B.     Manual  of  Blow-pipe  Analysis 8vo,  *2  50 

Courtney,  C.  F.     Masonry  Dams 8vo,  3  50 

Co  well,  W.  B.     Pure  Air,  Ozone,  and  Water i2mo,  *2  00 

Craig,  T.     Motion  of  a  Solid  in  a  Fuel.     (Science  Series  No.  49.) 

i6mo,  o  50 

Wave  and  Vortex  Motion. '    (Science  Series  No.  43.) .  i6mo,  o  50 

Cramp,  W.     Continuous  Current  Machine  Design 8vo,  *2  50 

Crocker,  F.  B.     Electric  Lighting.     Two  Volumes.     8vo. 

Vol.    I.     The  Generating  Plant 3  00 

Vol.  II.     Distributing  Systems  and  Lamps 3  00 

Crocker,  F.  B.,  and  Arendt,  M.     Electric  Motors 8vo,  *2  50 

Crocker,  F.  B.,  and  Wheeler,  S.  S.     The  Management  of  Electri- 
cal Machinery 1 2mo,  *  1  00 

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Its  Applications.     (Westminster  Series.) 8vo,  *2  00 

Crosskey,  L.  R.     Elementary  Prospective 8vo,  1  00 

Crosskey,  L.  R.,  and  Thaw,  J.     Advanced  Perspective 8vo,  1  50 

Culley,  J.  L.     Theory  of  Arches.     (Science  S~:hs  Ho.  87.).  16 mo,  o  50 


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Davies,  D.  C.     Metalliferous  Minerals  and  Mining 8vo,  5  00 

Earthy  Minerals  and  Mining 8vo,  5  00 

Davies,  E.  H.     Machinery  for  Metalliferous  Mines 8vo,  8  00 

Davies,  F.  H.      Electric  Power  and  Traction 8vo,  *2  00 

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Dadourian,  H.  M.      Analytical  Mechanics {In  Press.) 

Dawson,  P.     Electric  Traction  on  Railways 8vo,  *g  00 

Day,  C.     The  Indicator  and  Its  Diagrams nmo,  *2  00 

Deerr,  N.     Sugar  and  the  Sugar  Cane 8vo,  *8  00 

Deite,  C.     Manual  of  Soapmaking.     Trans,  by  S.  T.  King.  -4to,  *5  00 
De  la  Coux,  H.     The  Industrial  Uses  of  Water.     Trans,  by  A. 

Morris 8vo,  *4  50 

Del  Mar,  W.  A.     Electric  Power  Conductors 8vo,  *2  00 

Denny,  G.  A.     Deep-Level  Mines  of  the  Rand 4to,  *io  00 

Diamond  Drilling  f.or  Gold *5  00 

De  Roos,  J.  D.  C.     Linkages.     (Science  Series  No.  47.). .  .i6mo,  050 

Derr,  W.  L.     Block  Signal  Operation Oblong  nmo,  *i  50 

Maintenance  of  Way  Engineering {In  Preparation.) 

Desaint,   A.     Three   Hundred  Shades  and  How  to  Mix  Them. 

8vo,  *io  00 

De  Varona,  A.     Sewer  Gases.     (Science  Series  No.  55.)...  i6mo,  o  50 
Devey,  R.  G.     Mill  and  Factory- Wiring.     (Installation  Manuals 

Series.) nmo,  *i  00 

Dibdin,  W.  J.     Public  Lighting  by  Gas  and  Electricity 8vo,  *8  00 

Purification  of  Sewage  and  Water 8vo,  6  50 

Dichman,  C.    Basic  Open-Hearth  Steel  Process 8vo,  *3  50 

Dietrich,  K.     Analysis  of  Resins,  Balsams,  and  Gum  Resins  .8vo,  *3  00 
Dinger,  Lieut.  H.  C.     Care  and  Operation  of  Naval  Machinery 

i2mo.  *2  00 
Dixon,  D.  B.     Machinist's  and  Steam  Engineer's  Practical  Cal- 
culator   i6mo,  mor.,  1  25 

Doble,  W.  A.    Power  Plant  Construction  on  the  Pacific  Coast.  {In  Press.) 
Dodd,  G.     Dictionary  of  Manufactures,  Mining,  Machinery,  and 

the  Industrial  Arts i2mo,  1  50 

Dorr,  B.  F.     The  Surveyor's  Guide  and  Pocket  Table-book. 

i6mo,  mor.,  2  00 

Down,  P  B.     Handy  Copper  Wire  Table i6mo,  *i  00 


12  D.  VAN  XOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Draper,    C.    H.     Elementary    Text-book    of    Light,    Heat   and 

Sound 1 2mo,  i  oo 

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Duckwall,  E.  W.  Canning  and  Preserving  of  Food  Products. 8 vo,  *5  00 
Dumesny,  P.,  and  Noyer,  J.     Wood  Products,  Distillates,  and 

Extracts 8vo,  *4  50 

Duncan,  W.  G.,  and  Penman,  D.     The  Electrical  Equipment  of 

Collieries 8vo,  *4  00 

Dunstan,  A.  E.,  and  Thole,  F.  T.  B.     Textbook  of  Practical 

Chemistry nmo,  *i  40 

Duthie,    A.     L.     Decorative    Glass    Processes.     (Westminster 

Series) 8vo,  *2  00 

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Dyson,  S.  S.     Practical  Testing  of  Raw  Materials 8vo,  *5  00 

Dyson,  S.  S.  and  Clarkson,  S.  S.     Chemical  Works 8vo,  *7  50 

Eccles,  R.G.,  and  Duckwall, E.W.   Food  Preservatives.  8 vo,  paper,  o  50 

Eddy,  H.  T.     Researches  in  Graphical  Statics 8vo,  1   50 

Maximum  Stresses  under  Concentrated  Loads 8vo,  1  50 

Edgcumbe,  K.     Industrial  Electrical  Measuring  Instruments. 

8vo,  *2  50 

Eissler,  M.     The  Metallurgy  of  Gold 8vo,  7  50 

The  Hydrometallurgy  of  Copper 8vo,  *4  50 

The  Metallurgy  of  Silver ■. 8 vo,  4  00 

The  Metallurgy  of  Argentiferous  Lead 8vo,  5  00 

Cyanide  Process  for  the  Extraction  of  Gold 8vo,  3  00 

A  Handbook  of  Modern  Explosives 8vo,  5  00 

Ekin,  T.   C.      Water   Pipe   and    Sewage    Discharge   Diagrams 

folio,  *3  00 
Eliot,  C.  W.,  and  Storer,  F.  H.    Compendious  Manual  of  Qualita- 
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Elliot,  Major  G.  H.     European  Light-house  Systems 8vo,  5  00 

Ennis,  Wm.  D.     Linseed  Oil  and  Other  Seed  Oils   8vo,  *4  00 

Applied  Thermodynamics 8vo,  *4   50 

Flying  Machines  To-day i2mo,  *i  50 

Vapors  for  Heat  Engines i2mo,  *i  00 

Erfurt,  J.     Dyeing  of  Paper  Pulp.     Trans,  by  J.  Hubner. .  .8vo,  *7  50 

Ermen,  W.  F.  A.     Materials  Used  in  Sizing i2mo,  *2  00 

Erskine -Murray,  J.     A  Handbook  of  Wireless  Telegraphy.  .8 vo,  *3  50 


D.  VAN  NOSTRAND  COMPANY  S  SHORT-TITLE  CATALOG     13 

Evans,  C.  A.     Macadamized  Roads (In  Press.) 

Ewing,  A.  J.     Magnetic  Induction  in  Iron 8vo,  *4  oo 

Fairie,  J.     Notes  on  Lead  Ores i2mo,  *i  oo 

Notes  on  Pottery  Clays i2mo,  *i  50 

Fairley,  W.,  and  Andre,  Geo.  J.     Ventilation  of  Coal  Mines. 

(Science  Series  No.  58.) i6mo,  o  50 

Fairweather,  W.  C.     Foreign  and  Colonial  Patent  Laws  . .  .8vo,  *3  00 

Fanning,  T.  T.     Hydraulic  and  Water-supply  Engineering. 8vo,  *5  00 
Fauth,   P.     The   Moon  in  Modern   Astronomy.     Trans,    by  J. 

McCabe 8vo,  *2  00 

Fay,  I.  W.     The  Coal-tar  Colors 8vo,  *4  00 

Fernbach,  R.  L.     Glue  and  Gelatine. 8vo,  *3  00 

Chemical  Aspects  of  Silk  Manufacture nmo,  *i  00 

Fischer,  E.     The  Preparation  of  Organic  Compounds.     Trans. 

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Fish,  J.  C.  L.     Lettering  of  Working  Drawings Oblong  80,  1  00 

Fisher,  H.  K.  C,  and  Darby,  W.  C.     Submarine  Cable  Testing. 

8vo,  *3  50 

Fiske,  Lieut.  B.  A«     Electricity  in  Theory  and  Practice 8vo,  2  50 

Fleischmann,  W.     The  Book  of  the  Dairy.     Trans,  by  C.  M. 

Aikman  8vo,  4  00 

Fleming,    J.    A.     The    Alternate-current    Transformer.     Two 

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Fleming,  J,  A.     Centenary  of  the  Electrical  Current 8vo,  *o  50 

Electric  Lamps  and  Electric  Lighting 8vo,  *3  00 

Electric  Laboratory  Notes  and  Forms 4to,  *5  00 

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Room.     Two  Volumes 8vo,  each,  *5  00 

Fleury,  P.     White  Zinc  Paints 12010,  *2  50 

Fluery,   H.     The  Calculus  Without    Limits  or    Infinitesimals. 

Trans,  by  C.  0.  Mailloux (In  Press.) 

Flynn,  P.  J.     Flow  of  Water.     (Science  Series  No.  84.). . .  i6mo,  o  50 

Hydraulic  Tables.     (Science  Series  No.  66.) i6mo,  o  50 

Foley,  N.     British  and  American  Customary  and  Metric  Meas- 
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Foster,    H.     A.     Electrical    Engineers'     Pocket-book.     (Sixth 

Edition.) i2mo,  leather,  5  00 

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Foster,    Gen.    J.    G.     Submarine    Blasting    in    Boston    (Mass.) 

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Fox,  W.  G.     Transition  Curves.     (Science  Series  No.  no.). i6mo,  o  50 
Fox,  W.,  and  Thomas,  C.  W.     Practical  Course  in  Mechanical 

Drawing i2mo,  1  25 

Foye,  J.  C.     Chemical  Problems.     (Science  Series  No.  60.). i6mo,  o  50 

Handbook    of    Mineralogy.       (Science     Series    No.    86.). 

i6mo,  o  50 

Francis,  J.  B.     Lowell  Hydraulic  Experiments 4to,  15  00 

Freudemacher,   P.   W.     Electrical   Mining  Installations.     (In- 
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Frith,  J.     Alternating  Current  Design 8vo,  *2  00 

Fritsch,  J.     Manufacture  of  Chemical  Manures.     Trans,  by 

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Frye,  A.  I.     Civil  Engineers'  Pocket-book ....  i2mo,  leather,  (In  Press.) 
Fuller,  G.  W.     Investigations  into  the  Purification  of  the  Ohio 

River 4to,  *io  00 

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Gairdner,  J.  W.  I.     Earthwork 8vo  (In  Press.) 

Gant,  L.  W.     Elements  of  Electric  Traction 8vo,  *2  50 

Garcia,  A.  J.  R.  V.     Spanish-English  Railway  Terms (In  Press.) 

Garforth,  W.  E.     Rules  for  Recovering  Coal  Mines  after  Explo- 
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Gaudard,  J.     Foundations.     (Science  Series  No.  34.) i6mo,  o  50 

Gear,  H.  B.,  and  Williams,  P.  F.     Electric  Central  Station  Dis- 
tributing Systems i2mo,  *3  00 

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Geikie,  J.     Structural  and  Field  Geology 8vo,  *4  00 

Gerber,   N.     Analysis  of  Milk,  Condensed  Milk,  and  Infants' 

M  Ik-Food 8vo,  1  25 

Gerhard,  W.  I .     Sanitation,  Water-supply  and  Sewage  Disposal 

of  Country  Houses nmo,  *a  00 

Gas  Lighting.     (Science  Series  No.  in.) i6mo,  o  50 


D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     15 

Gerhard,  W.  P.      Household  Wastes.     (Science  Series  No.  97.) 

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Sanitary  Drainage  of  Buildings.     (Science  Series  No.  93.) 

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Gerhardi,  C.  W.  H.     Electricity  Meters 8vo,  *4  00 

Geschwind,  L.     Manufacture  of  Alum  and  Sulphates.     Trans. 

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Gibbs,  W.  E.     Lighting  by  Acetylene i2mo,  *i  50 

Physics  of  Solids  and  Fluids.     (Carnegie  Technical  Schools 

Text-books.) *i   50 

Gibson,  A.  H.     Hydraulics  and  Its  Application 8vo,  *5  00 

Water  Hammer  in  Hydraulic  Pipe  Lines i2mo,  *2  00 

Gilbreth,  F.  B.     Motion  Study.     A  Method  for  Increasing  the 

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Gilbreth,  F.  B.     Primer  of  Scientific  Management i2mo,  *i  00 

Gillmore,  Gen.  Q.  A.     Limes,  Hydraulic  Cements  and  Mortars. 

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Golding,  H.  A.     The  Theta-Phi  Diagram i2mo,  *i   25 

Goldschmidt,  R.     Alternating  Current  Commutator  Motbr  .8vo,  *3  00 

Goodchild,  W.     Precious  Stones.     (Westminster  Series.) ..  .8vo,  *2  00 

Goodeve,  T.  M.     Textbook  on  the  Steam-engine i2mo,  2  00 

Gore,  G.     Electrolytic  Separation  of  Metals 8vo,  *3  50 

Gould,  E.  S.     Arithmetic  of  the  Steam-engine i2mo,  1  00 

Calculus.     (Science  Series  No.  112.) i6mo,  o  50 

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Gray,  J.     Electrical  Influence  Machines i2mo,  2  00 

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Greenhill,  G.     Dynamics  of  Mechanical  Flight 8vo,  *2  50 

Greenwood,  E.     Classified  Guide  to  Technical  and  Commercial 

Books 8vo,  *3  00 

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Griffiths,  A.  B.     A  Treatise  on  Manures i2mo,  3  00 


16    D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Griffiths,  A.  B.     Dental  Metallurgy 8vo,  *3  50 

Gross,  E.     Hops 8vo,  *4  50 

Grossman,  J.     Ammonia  and  its  Compounds nmo,  *i  25 

Groth,  L.  A.     Welding  and  Cutting  Metals  by  Gases  or  Electric- 
ity   8vo,  *3  00 

Grover,  F.     Modern  Gas  and  Oil  Engines 8vo,  *2  00 

Gruner,  A.     Power-loom  Weaving 8vo,  *3  00 

Guldner,    Hugo.      Internal-Combustion    Engines.      Trans,    by 

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Gunther,  C.  0.     Integration nmo,  *i  25 

Gurden,  R.  L.     Traverse  Tables folio,  half  mor.  *7  50 

Guy,  A.  E.     Experiments  on  the  Flexure  of  Beams 8vo,  *i  25 

Haeder,  H.     Handbook  on  the  Steam-engine.     Trans,  by  H.  H. 

P.  Powles nmo,  3  00 

Haenig,  A.     Emery  and  the  Emery  Industry i2mo,  *2  50 

Hainbach,  R.     Pottery  Decoration.     Trans,  by  C.  Slater.  .  nmo,  *3  00 

Hale,  W.  J.     Calculations  of  General  Chemistry. ., nmo,  *i  00 

Hall,  C.  H.     Chemistry  of  Paints  and  Paint  Vehicles nmo,  *2  00 

Hall,  R.  H.     Governors  and  Governing  Mechanism. nmo,  *2  00 

Hall,  W.  S.     Elements  of  the  Differential  and  Integral  Calculus 

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Halsey,  F.  A.     Slide  Valve  Gears nmo,  1 

The  Use  of  the  Slide  Rule.   (Science  Series.) i6mo,  0 

Worm  and  Spiral  Gearing.     (Science  Scries.) i6mo,  o 

Hamilton,  W.  G.     Useful  Information  for  Railway  Men. .  i6mo,  1 
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Hancock,  H.     Textbook  of  Mechanics  and  Hydrostatics 8vo,  1 

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Haskins,  C.  H.     The  Galvanometer  and  Its  Uses i6mo,  1 

Hatt,  J.  A.  H.     The  Colorist square  nmo,  *i 


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8vo, 

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Hering,  D.  W.     Essentials  of  Physics  for  College  Students. 

8vo,     *i  75 
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*2 

00 

*5 

00 

*3 

00 

*5 

00 

*5 

00 

*2 

50 

*2 

50 

*3 

5o 

*2 

50 

2 

00 

*3 

50 

*5 

00 

*5 

00 

*3 

50 

*5 

00 

*i 

00 

*i 

00 

3 

00 

18 


Hildenbrand,  B.  W.     Cable-Making.      (Science  Series  No.  32.) 

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Hildich,  H.     Concise  History  of  Chemistry i2mo,  *i  25 

Hill,  J.  W.     The  Purification  of  Public  Water  Supplies.     New 

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Hobart,  J.  F.     Hard  Soldering,   Soft  Soldering,  and  Brazing. 

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Hoff,  J.  N.     Paint  and  Varnish  Facts  and  Formulas nmo,  *i  50 

Hoff,  Com.W.  B.  The  Avoidance  of  Collisions  at  Sea.  i6mo,  mor.,  0  75 

Hole,  W.     The  Distribution  of  Gas 8vo,  *7  50 

Holley,  A.  L.     Railway  Practice folio,  12  00 

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Houghton,  C.  E.     The  Elements  of  Mechanics  of  Materials,  nmo,  *2  00 

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D.  VAX  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG     29 

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Hurst,  G.  H.     Handbook  of  the  Theory  of  Color 8vo,  *2  50 

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Hurst,  H.  E.,  and  Lattey,  R.  T.     Text-book  of  Physics 8vo,  *3  00 

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Vol.    II.  Sound  and  Light 

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Hyde,  E.  W.     Skew  Arches.     (Science  Series  No.  15.)..  .  .  i6mo,  o  50 

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Ingle,  H.     Manual  of  Agricultural  Chemistry 8vo,  *3  00 

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20     D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Jacob,  A.,  and  Gould,  E.  S.     On  the  Designing  and  Construction 

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Jamieson,  A.     Text  Book  on  Steam  and  Steam  Engines. .  .  .  8vo,  3  00 

Elementary  Manual  on  Steam  and  the  Steam  Engine  .  12010,  1   50 

Jannettaz,  E.     Guide  to  the  Determination  of  Rocks.     Trans. 

by  G.  W.  Plympton.. , nmo,  1  50 

Jehl,  F.     Manufacture  of  Carbons 8vo,  *4  00 

Tennings,    A.    S.     Commercial    Paints    and    Painting.     (West- 
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Jennison,  F.  H.     The  Manufacture  of  Lake  Pigments 8vo,  *3  00 

Jepson,  G.     Cams  and  the  Principles  of  their  Construction. .  .8vo,  *i  50 

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Jockin,  W.     Arithmetic  of  the  Gold  and  Silversmith i2mo,  *i  00 

Johnson,  G.  L.     Photographic  Optics  and  Color  Photography. 

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Johnston,  J.  F.  W.,  and  Cameron,  C.     Elements  of  Agricultural 

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Jiiptner,  H.  F.  V.     Siderology:  The  Science  of  Iron 8vo,  *5  00 

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Kapp,  G.     Alternate  Current  Machinery.     (Science  Series  No. 

96.) i6mo,  o  50 

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Kelsey,   W.    R.       Continuous-current    Dynamos  and  Motors. 

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Hydrogen   Spectrum 8vo,  paper,  *o  50 

Kemp,  J.  F.     Handbook  of  Rocks 8vo,  *i  50 

Kendall,  E.     Twelve  Figure  Cipher  Code 4to,  *i5  00 

Kennedy,    A.    B.    W.,    and   Thurston,    R,    H.     Kinematics    of 

Machinery.     (Science  Series  No.  54.) i6mo,  o  50 

Kennedy,  A.  B.  W.,  Unwin,  W.  C,  and  Idell,  F.  E.     Compressed 

Air.     (Science  Series  No.  106.) i6mo,  o  50 

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Kennelly,  A.  E.     Electro-dynamic  Machinery. 8vo,  1  50 

Kent,  W.     Strength  of  Materials.     (Science  Series  No.  41.).  i6mo,  o  50 

Kershaw,  J.  B.  C.     Fuel,  Water  and  Gas  Analysis 8vo,  *2  50 

Electrometallurgy.     (Westminster  Series.) 8vo,  *2  00 

The  Electric  Furnace  in  Iron  and  Steel  Production. .  i2mo,  *i  50 

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Continuous  Current  Armatures 8vo,  *i  50 

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Kirkaldy,    W.    G.     David    Kirkaldy's    System    of    Mechanical 

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Klein,  J.  F.     Design  of  a  High  speed  Steam-engine 8vo,  *5  00 

Physical  Significance  of  Entropy 8vo,  *i  50 

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Knox,  W.  F.     Logarithm  Tables (In  Preparation.) 

Knott,  C.  G.,  and  Mackay,  J.  S.     Practical  Mathematics.  .  .  8vo,       2  00 

Koester,  F.     Steam-Electric  Power  Plants 4to, 

Hydroelectric  Developments  and  Engineering... .  4to, 

Koller,  T.     The  Utilization  of  Waste  Products 8vo, 

Cosmetics 8vo, 

Kretchmar,  K.     Yarn  and  Warp  Sizing 8vo, 


*5 

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*5 

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*3 

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*2 

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*4 

00 

*3 

5^ 

*3 

00 

*3 

00 

*7 

50 

■6 

00 

1 

25 

0 

60 

0 

50 

Lambert,  T.     Lead  and  its  Compounds 8vo, 

Bone  Products  and  Manures 8vo, 

Lamborn,  L.  L.     Cottonseed  Products 8vo, 

Modern  Soaps,  Candles,  and  Glycerin 8vo, 

Lamprecht,  R.     Recovery  Work  After  Pit  Fires.      Trans,   by 

C.  Salter 8vo,     *4  00 

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Lamer,  E.  T.     Principles  of  Alternating  Currents i2mo, 

Larrabee,    C.    S.     Cipher   and   Secret   Letter   and   Telegraphic 

Code i6mo, 

La  Rue,  B.  F.     Swing  Bridges.      (Science  Series  No.  107.) .  i6mo, 
Lassar-Cohn,  Dr.     Modern  Scientific  Chemistry.     Trans,  by  M. 

M.  Pattison  Muir nmo,     *2  00 

Latimer,  L.  H.,  Field,  C.  J.,  and  Howell,  J.  W.     Incandescent 

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Latta,  M.  N.     Handbook  of  American  Gas-Engineering  Practice. 

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Leask,  A.  R.     Breakdowns  at  Sea i2mo,       2  00 

Refrigerating  Machinery i2mo,       2  00 

Lecky,  S.  T.  S.     "  Wrinkles  "  in  Practical  Navigation  ....  8vo,     *8  00 
Le  Doux,  M.     Ice-Making  Machines.     (Science  Series  No.  46.) 

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Leeds,  C.  C.    Mechanical  Drawing  for  Trade  Schools .  oblong,  4to, 

High  School  Edition *i  25 

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Lemstrom,  S.     Electricity  in  Agriculture  and  Horticulture.  .8 vo,     *i  50 
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Lewes,  V.  B.     Liquid  and  Gaseous  Fuels.     (Westminster  Series.) 

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Lieber,  B.  F.     Lieber's  Standard  Telegraphic  Code  . 8vo,  *io  00 

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Locke,  A.  G.  and  C.  G.     Manufacture  of  Sulphuric  Acid 8vo,     10  00 

Lockwood,  T.  D.     Electricity,  Magnetism,  and  Electro-teleg- 
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Lodge,  O.  J.     Elementary  Mechanics. i2mo,       1  50 

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Loring,  A.  E.     A  Handbook  of  the  Electromagnetic  Telegraph. 

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Lubschez,  B.  J.    Perspective 121310,     *i  50 

Lucke,  C.  E.     Gas  Engine  Design 8vo,     *3  00 

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2  vols (In  Preparation.) 


24     D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

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Macewen,  H.  A.     Food  Inspection 8vo,  *2  50 

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Mackie,  J.     How  to  Make  a  Woolen  Mill  Pay 8vo,  *2  00 

Mackrow,  C.  Naval  Architect's  and  Shipbuilder's  Pocket- 
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Maguire,  Wm.  R.     Domestic  Sanitary  Drainage  and  Plumbing 

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(Science  Series  No.  10.) i6mo, 

Mansfield,  A.  N.     Electro-magnets.     (Science  Series  No.  64) 

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Construction  and  Working  of  Pumps i2mo,  *i  50 

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Marlow,  T.  G.     Drying  Machinery  and  Practice. 8vo,  *5  00 

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Marshall,  W.J.,  and  Sankey,  H.  R.    Gas  Engines.     (Westminster 

Series.) 8vo,  *2  00 

Martin,  G.     Triumphs  and  Wonders  of  Modern  Chemistry. 

8vo,  *2  00 

Martin,  N.     Reinforced  Concrete 8vo,  *2  50 

Massie,  W.  W.,  and  Underhill,  C.  R.     Wireless  Telegraphy  and 

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Matheson,  D.     Australian  Saw-Miller's  Log  and  Timber  Ready 

Reckoner nmo,  leather,  1  50 

Mathot,  R.  E.     Internal  Combustion  Engines 8vo,  *6  00 

Maurice,  W.     Electric  Blasting  Apparatus  and  Explosives  ..8vo,  *3  50 

Shot  Firer's  Guide 8vo,  *  1  50 

Maxwell,  J.  C.     Matter  and  Motion.     (Science  Series  No.  36.) 

i6mo,  o  50 
Maxwell,  W.  H.,  and  Brown,  J.  T.     Encyclopedia  of  Municipal 

and  Sanitary  Engineering 4to,  *io  00 

Mayer,  A.  M.     Lecture  Notes  on  Physics 8vo,  2  00 

McCullough,  R.  S.     Mechanical  Theory  of  Heat .  8vo,  3  50 

Mcintosh,  J.  G.     Technology  of  Sugar 8vo,  *4  50 

Industrial  Alcohol 8vo,  *3  00 

Manufacture  of  Varnishes  and  Kindred  Industries.    Three 

Volumes.     8vo. 

Vol.  I.     Oil  Crushing,  Refining  and  Boiling *3  50 

Vol.  II.     Varnish  Materials  and  Oil  Varnish  Making *4  00 

Vol.  III.     Spirit  Varnishes  and  Materials *4  5© 

McKnight,   J.   D.,   and   Brown,    A.   W.     Marine   Multitubular 

Boilers *i  50 

McMaster,  J.  B.     Bridge  and  Tunnel  Centres.     (Science  Series 

No.  20.) i6mo,  o  50 

McMechen,  F.  L.     Tests  for  Ores,  Minerals  and  Metals..  .  i2mo,  *i  00 

McNeill,  B.     McNeill's  Code 8vo,  *6  00 

McPherson,  J.  A.     Water-works  Distribution 8vo,  2  50 

Melick,  C.  W.     Dairy  Laboratory  Guide 121110,  *i  25 

Merck,  E.     Chemical  Reagents ;  Their  Purity  and  Tests..  .  .8vo,  *i  50 


26     D.  VAX  NOSTRAND  COMPANY'S  SHORT  TITLP]  CATALOG 

Merritt,  Wm.  H.  Field  Testing  for  Gold  and  Silver .  i6mo,  leather,       i  50 

Messer,  W.  A.     Railway  Permanent  Way 8vo    (In  Press.) 

Meyer,  J.  G.  A.,  and  Pecker,  C.  G.     Mechanical  Drawing  and 

Machine  Design 4to,       5  00 

Mitchell,  C.  F.,  and  G.  A.    Building  Construction  and  Drawing. 

Elementary i2mc,     *i  50 

Advanced i2mo,     *2  50 

Michell,  S.     Mine  Drainage 8vo,     10  00 

Mierzinski,  S.     Waterproofing  of  Fabrics.     Trans,  by  A.  Morris 

and  H.  Robson 8vo,     *2  50 

Miller,  E.  H.     Quantitative  Analysis  for  Mining  Engineers ..  8vo,     *i  50 
Miller,  G.  A.     Determinants.     (Science  Series  No.  105.).  .i6mo, 

Milroy,  M.  E.  W.     Home  Lace -making i2mo,     *i  00 

Minifie,  W.     Mechanical  Drawing 8vo,     *4  00 

Mitchell,  C.  A.,  and  Prideaux,  R.  M.     Fibres  Used  in  Textile  and 

Allied  Industries 8vo,     *3  00 

Modern  Meteorology i2mo,       1  50 

Monckton,  C.  C.  F.     Radiotelegraphy.     (Westminster  Series.) 

8vo,     *2  00 
Monteverde,  R.  D.     Vest  Pocket  Glossary  of  English-Spanish, 

Spanish-English  Technical  Terms 64mo,  leather,     *i  00 

Moore,  E.  C.  S.     New  Tables  for  the  Complete  Solution  of 

Ganguillet  and  Kutter's  Formula 8vo,     *5  00 

Morecroft,  J.  H.,  and  Hehre,  F.  W.    Testing  Electrical  Ma- 
chinery   8vo,     *i  50 

Moreing,  C.  A.,  and  Neal,  T.     New  General  and  Mining  Tele- 
graph Code 8vo,     *5  00 

Morgan,  A.  P.     Wireless  Telegraph  Construction  for  Amateurs. 

i2mo,     *i  50 

Moses,  A.  J.     The  Characters  of  Crystals 8vo,     *2  00 

Moses,  A.  J.,  and  Parsons,  C.  I.     Elements  of  Mineralogy ..  8vo,     *2  50 
Moss,    S.    A.     Elements    of    Gas    Engine    Design.     (Science 

Series.) i6mo,       o  50 

The  Lay-out  of  Corliss  Valve  Gears.      (Science  Series) . 

i6mo,      o  50 

Mulford,  A.  C.     Boundaries  and  Landmarks 8vo,     *i  00 

Mullin,  J.  P.     Modern  Moulding  and  Pattern-making.  .  .  .  nmo,       2  50 
Munby,  A.  E.     Chemistry  and  Physics  of  Building  Materials. 

(Westminster  Series.) 8vo,     *2  00 


D.  VAN  NOSTRAND  COMPANY'S  SHOUT  TITLE  CATALOG      27 

Murphy,  J.  G.     Practical  Mining • i6mo,  i  oo 

Murray,  J.  A.     Soils  and  Manures.     (Westminster  Series.)  8vo,  *2  oo 

Naquet,  A.     Legal  Chemistry i2mo,  2  00 

Nasmith,  J.     The  Student's  Cotton  Spinning 8vo,  3  00 

Recent  Cotton  Mill  Construction 12010,  2   00 

Neave,  G.  B.,  and  Heilbron,  I.  M.     Identification  of  Organic 

Compounds nmo,  *i  25 

Neilson,  R.  M.     Aeroplane  Patents 8vo,  *2  00 

Nerz,  F.     Searchlights.     Trans,  by  C.  Rodgers 8vo,  *3  00 

Nesbit,  A.  F.     Electricity  and  Magnetism (In  Preparation.) 

Neuberger,   H.,   and   Noalhat,  H.     technology   of   Petroleum. 

Trans,  by  J.  G.  Mcintosh. 8vo,  *io  00 

Newall,  J.  W.     Drawing,  Sizing  and  Cutting  Bevel-gears.  .8vo,  1  50 

Nicol,  G.     Ship  Construction  and  Calculations 8vo,  *4  50 

Nipher,  F.  E.     Theory  of  Magnetic  Measurements i2mo,  1  00 

Nisbet,  H.     Grammar  of  Textile  Design 8vo,  *3  00 

Nolan,  H.     The  Telescope.     (Science  Series  No.  51.) i6mo,  o  50 

Noll,  A.     How  to  Wire  Buildings i2mo,  1  50 

Nugent,  E.     Treatise  on  Optics i2mo,  1  50 

O'Connor,  H.     The  Gas  Engineer's  Pocketbook.  ..  i2mo,  leather,  3  50 

Petrol  Air  Gas i2mo,  *o  75 

Ohm,  G.  S.,  and  Lockwood,  T.  D.     Galvanic  Circuit.     Trans,  by 

William  Francis.     (Science  Series  No.  102.).  .  .  .  i6mo,  o  50 

Olsen,  J.  C.  Text  book  of  Quantitative  Chemical  Analysis .  .8vo,  *4  00 
Olsson,  A.     Motor  Control,  in  Turret  Turning  and  Gun  Elevating. 

(U.  S.  Navy  Electrical  Series,  No.  1.)  .  ...nmo,  paper,  *o  50 

Oudin,  M.  A.     Standard  Polyphase  Apparatus  and  Systems  .  .8vo,  *3  00 

Pakes,  W.  C.  C,  and  Nankivell,  A.  T.    The  Science  of  Hygiene. 

8vo,  *i  75 
Palaz,  A.     Industrial  Photometry.     Trans,  by  G.  W.  Patterson, 

Jr 8vo,  *4  00 

Pamely,  C.     Colliery  Manager's  Handbook 8vo,  *io  00 

Parr,  G.  D.  A.     Electrical  Engineering  Measuring  Instruments. 

8vo,  *3  50 
Parry,  E.  J.     Chemistry  of  Essential  Oils  and  Artificial  Per- 
fumes  8vo,  *5  00 


28      D.  VAN  NOSTRAND  COMPANY  S  SHORT  TITLE  CATALOG 

Parry,  E.  J.     Foods  and  Drugs.     Two  Volumes 8vo. 

Vol.    I.     Chemical  and  Microscopical  Analysis  of  Food 

and  Drugs *7 .  50 

Vol.  II.     Sale  of  Food  and  Drugs  Acts *3  00 

Parry,  E.  J.,  and  Coste,  J.  H.     Chemistry  of  Pigments 8vo,  *4  50 

Parry,  L.  A.     Risk  and  Dangers  of  Various  Occupations 8vo,  *3  00 

Parshall,  H.  F.,  and  Hobart,  H.  M.     Armature  Windings  ....  4to,  *7  50 

Electric  Railway  Engineering 4to,  *io  00 

Parshall,  H.  F.,  and  Parry,  E.     Electrical  Equipment  of  Tram- 
ways  (In  Press.) 

Parsons,  S.  J.     Malleable  Cast  Iron 8vo,  *2  50 

Partington,  J.  R.    Higher  Mathematics  for  Chemical  Students 

i2mo,  *2  00 

Passmore,  A.  C.     Technical  Terms  Used  in  Architecture  ...8vo,  *3  50 

Paterson,  G.  W.  L.     Wiring  Calculations i2mo,  *2  00 

Patterson,  D.     The  Color  Printing  of  Carpet  Yarns 8vo,  *3  50 

Color  Matching  on  Textiles 8vo,  *3  00 

The  Science  of  Color  Mixing 8vo,  *3  00 

Paulding,  C.  P.     Condensation  of  Steam  in  Covered  and  Bare 

Pipes 8vo,  *2  00 

Paulding.  C.  P.     Transmission  of  Heat  through  Cold-storage 

Insulation i2mo,  *i  00 

Payne,  D.  W.     Iron  Founders'  Handbook (In  Press.) 

Peddie,  R.  A.    Engineering  and  Metallurgical  Books. .  .  .  i2mo,  *i  50 

Peirce,  B.     System  of  Analytic  Mechanics 4to,  10  00 

Pendred,  V.     The  Railway  Locomotive.     (Westminster  Series.) 

8vo,  *2  00 

Perkin,  F.  M.     Practical  Method  of  Inorganic  Chemistry ..  i2mo,  *i  00 

Perrigo,  0.  E.     Change  Gear  Devices 8vo,  1  00 

Perrine,  F.  A.  C.     Conductors  for  Electrical  Distribution  .  .  .  8vo,  *3  50 

Perry,  J.     Applied  Mechanics 8vo,  *2  50 

Petit,  G.     White  Lead  and  Zinc  White  Paints 8vo,  *i  50 

Petit,   R.     How  to  Build  an  Aeroplane.     Trans,   by  T.   O'B. 

Hubbard,  and  J.  H.  Ledeboer 8vo,  *i  50 

Pettit,  Lieut.  J.  S.     Graphic  Processes.     (Science  Series  No.  76.) 

i6mo,  o  50 
Philbrick,  P.  H.     Beams  and  Girders.     (Science  Series  No.  88.) 

i6mo, 

Phil   ps,  J.     Engineering  Chemistry 8vo,  *4  50 


D.  VAN  .NOSTRA XI >  COMPANY'S  SHORT-TITLE  CATALOG  20 

Phillips,  J.     Gold  Assaying 8vo,  *2  50 

Dangerous  Goods 8vo,  3  50 

Phin,  J.     Seven  Follies  of  Science nmo,  *i  25 

Pickworth,  C.  N.     The  Indicator  Handbook.     Two  Volumes 

nmo,  each,  1  50 

Logarithms  for  Beginners i2mo,  boards,  o  50 

The  Slide  Rule i2mo,  1  00 

Plattner's  Manual  of    Blowpipe  Analysis.     Eighth  Edition,  re- 
vised.    Trans,  by  H.  B.  Cornwall 8vo,  *4  00 

Plympton,  G.W.  The  Aneroid  Barometer.  (Science  Series.). i6mo,  o  50 

How  to  become  an  Engineer.     (Science  Series  No.   100.) 

i6mo,  o  50 
Plympton,  G.  W.     Van  Nostrand's  Table  Book.     (Science  Series 

No.  104.) i6mo,  o  50 

Pochet,  M.  L.     Steam  Injectors.     Translated  from  the  French. 

(Science  Series  No.  29.) i6mo,  0  50 

Pocket  Logarithms  to  Four  Places.     (Science  Series.) i6mo,  o  50 

leather,  1  00 

Polleyn,  F.     Dressings  and  Finishings  for  Textile  Fabrics .  8vo,  *3  00 

Pope,  F.  L.     Modern  Practice  of  the  Electric  Telegraph..  .  .  8vo,  1  50 
Popplewell,   W.   C.     Elementary  Treatise    on   Heat   and   Heat 

Engines t i2mo,  *3  00 

Prevention  of  Smoke 8vo,  *3  50 

— —  Strength  of  Minerals 8vo,  *i  75 

Porter,  J.  R.     Helicopter  Flying  Machines 12010,  1  25 

Potter,  T.     Concrete 8vo,  *3  00 

Potts,  H.  E.  Chemistry  of  the  Rubber  Industry.     (Outlines  of 

Industrial  Chemistry.) 8vo,  *2  00 

Practical  Compounding  of  Oils,  Tallow  and  Grease 8vo,  *3  50 

Practical  Iron  Founding i2mo,  1  50 

Pratt,  K.     Boiler  Draught nmo,  *i  25 

Pray,  T.,  Jr.     Twenty  Years  with  the  Indicator 8vo,  2  50 

Steam  Tables  and  Engine  Constant 8vo,  2  00 

Calorimeter  Tables 8vo,  1  00 

Preece,  W.  H.     Electric  Lamps. (In  Press.) 

Prelini,  C.     Earth  and  Rock  Excavation. 8vo,  *3  00 

Dredges  and  Dredging 8vo,  *3  00 

Graphical  Determination  of  Earth  Slopes 8vo,  *2  00 

Tunneling 8vo,  *3  00 


30   D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG 

Prescott,  A.  B.     Organic  Analysis 8vo,  5  00 

Prescott,  A.    B.,  and   Johnson,    0.    C.     Quantitative   Chemical 

Analysis 8vo,  *3  50 

Prescott,  A.  B.,  and  Sullivan,  E.  C.     First  Book  in  Qualitative 

Chemistry i2mo,  *i  50 

Prideaux,  E.  B.  R.     Problems  in  Physical  Chemistry 8vo,  *2  00 

Pritchard,  0.  G.     The  Manufacture  of  Electric-light  Carbons. 

8vo,  paper,  *o  60 
Pullen,  W.  W.  F.     Application  of  Graphic  Methods  to  the  Design 

of  Structures i2mo,  *2  50 

Injectors:  Theory,  Construction  and  Working i2mo,  *i  50 

Pulsifer,  W.  H.     Notes  for  a  History  of  Lead 8vo,  4  00 

Purchase,  W.  R.     Masonry i2mo,  *3  00 

Putsch,  A.     Gas  and  Coal-dust  Firing 8vo,  *3  00 

Pynchon,  T.  R.     Introduction  to  Chemical  Physics 8vo,  3  00 

Rafter,  G.  W.     Mechanics  of  Ventilation.     (Science  Series  No. 

330 i6mo,  o  50 

Potable  Water.     (Science  Series  No.  103.) i6mo,  o  50 

Treatment  of  Septic  Sewage.     (Science  Series.) ....  i6mo,  o  50 

Rafter,  G.  W.,  and  Baker,  M.  N.     Sewage  Disposal  in  the  United 

States • 4to,  *6  00 

Raikes,  H.  P.     Sewage  Disposal  Works 8vo,  *4  00 

Railway  Shop  Up-to-Date 4to,  2  00 

Ramp,  H.  M.     Foundry  Practice (In  Press.) 

Randall,  P.  M.     Quartz  Operator's  Handbook i2mo,  2  00 

Randau,  P.     Enamels  and  Enamelling 8vo,  *4  00 

Rankine,  W.  J.  M.     Applied  Mechanics 8vo,  5  00 

Civil  Engineering. 8vo,  6  50 

Machinery  and  Millwork. 8vo,  5  00 

Rankine,   W.   J.    M.     The   Steam-engine   and   Other  Prime 

Movers 8vo,  5  00 

Useful  Rules  and  Tables 8vo,  4  00 

Rankine,  W.  J.  M.,  and  Bamber,  E.  F.     A  Mechanical  Text- 
book  8vo,  3  50 

Raphael,  F.  C.     Localization  of    Faults  in  Electric  Light  and 

Power  Mains 8vo,  *3  00 

Rasch,  E.     Electric  Arc  Phenomena.     Trans,  by  K.  Tornberg. 

(In  Press.) 


*2 

50 

*2 

50 

*3 

50 

*4 

50 

D.  VAX  NOSTRAND  COMPANY'S  SHOIIT-TITLK  CATALOG     3J 

Rathbone,  R.  L.  B.     Simple  Jewellery 8vo,     *2  oo 

Rateau,    A.     Flow  of   Steam   through   Nozzles    and    Orifices. 

Trans,  by  H.  B.  Brydon 8vo,     *i   50 

Rausenberger,  F.     The  Theory  of  the  Recoil  of  Guns. ....  8vo,     *4  50 
Rautenstrauch,  W.     Notes  on  the  Elements  of  Machine  Design, 

8vo,  boards,     *i  50 
Rautenstrauch,  W.,  and  Williams,  J.  T.     Machine  Drafting  and 

Empirical  Design. 

Part    I.  Machine  Drafting 8vo,     *i  25 

Part  II.  Empirical  Design (In  Preparation.) 

Raymond,  E.  B.     Alternating  Current  Engineering i2mo, 

Rayner,  H.     Silk  Throwing  and  Waste  Silk  Spinning 8vo, 

Recipes  for  the  Color,  Paint,  Varnish,  Oil,  Soap  and  Drysaltery 

Trades 8vo, 

Recipes  for  Flint  Glass  Making nmo, 

Redfern,    J.    B.     Bells,    Telephones.     (Installation    Manuals 

Series.) i6mo  (In  Press.) 

Redwood,  B.     Petroleum.     (Science  Series  No.  92.) i6mo,       o  50 

Reed,  T.     Guide  to  the  Slide  Rule (In  Press.) 

Reed's  Engineers'  Handbook 8vo, 

Key  to  the  Nineteenth  Edition  of  Reed's  Engineers'  Hand- 
book  8vo, 

Reed's  Useful  Hints  to  Sea-going  Engineers nmo, 

Marine  Boilers i2mo, 

Reinhardt,   C.   W.     Lettering   for   Draftsmen,   Engineers,   and 

Students oblong  4to,  boards, 

The  Technic  of  Mechanical  Drafting. .  .  oblong  4to,  boards, 

Reiser,  F.     Hardening  and  Tempering  of  Steel.     Trans,  by  A. 

Morris  and  H.  Robson i2mo,     *2  50 

Reiser,  N.     Faults  in  the  Manufacture  of  Woolen  Goods.     Trans. 

by  A.  Morris  and  H.  Robson 8vo, 

Spinning  and  Weaving  Calculations 8vo, 

Renwick,  W.  G.     Marble  and  Marble  Working 8vo, 

Reynolds,    0.,    and   Idell,    F.    E.     Triple    Expansion    Engines. 

(Science  Series  No.  99.) i6mo, 

Rhead,  G.  F.     Simple  Structural  Woodwork i2mo, 

Rhead,  G.  W.     British  Pottery  Marks 8vo, 

Rice,  J.  M.,  and  Johnson,  W.  W.     A  New  Method  of  Obtaining 

the  Differential  of  Functions i2mo,       o  50 


*5 

00 

*3 

00 

1 

50 

2 

00 

1 

00 

*i 

00 

+  2 

50 

*5 

00 

5 

00 

0 

5c 

*  1 

00 

*3 

00 

32     D.  VAN  NOSTKAND  COMPANY'S  SHORT-TITLE  CATALOG 

Richards,  W.  A.  and  North,  H.  B.     Manual  of  Cement  Testing  *i  50 

Richardson,  J.     The  Modern  Steam  Engine 8vo,  *3  50 

Richardson,  S.  S.     Magnetism  and  Electricity i2mo,  *2  00 

Rideal,  S.     GJue  and  Glue  Testing 8vo,  *4  00 

Rimmer,  E.  J.     Boiler  Explosions 8vo,  *i  75 

Rings,  F.     Concrete  in  Theory  and  Practice i2mo,  *2  50 

Ripper,  W.     Course  of  Instruction  in  Machine  Drawing. .  .folio,  *6  00 
Roberts,  F.  C.     Figure  of  the  Earth.     (Science  Series  No.  79.) 

i6mo,  o  50 
Roberts,  J.,  Jr.     Laboratory  Work  in  Electrical  Engineering 

8vo,  *2  00 

Robertson,  L.  S.     Water-tube  Boilers 8vo,  *2  00 

Robinson,  J.  B.     Architectural  Composition 8vo,  *2  50 

Robinson,  S.  W.     Practical  Treatise  on  the  Teeth  of  Wheels. 

(Science  Series  No.  24.) i6mo,  o  50 

Railroad  Economics.     (Science  Series  No.  59.) i6mo,  o  50 

Wrought  Iron  Bridge  Members.     (Science  Series  No.  60.) 

i6mo,  o  50 

Robson,  J.  H,     Machine  Drawing  and  Sketching 8vo,  *i  50 

Roebling,  J.  A.     Long  and  Short  Span  Railway  Bridges.  .    folio,  2500 

Rogers,  A.     A  Laboratory  Guide  of  Industrial  Chemistry ..  i2mo,  *i  50 

Rogers,  A.,  and  Aubert,  A.  B.     Industrial  Chemistry 8vo,  *5  00 

Rogers,  F.     Magnetism  of  Iron  Vessels.     (Science  Series  No.  30.) 

i6mo,  o  50 
Rohland,  P.     Colloidal  and  its  Crystalloidal  State  of  Matter. 

Trans,  by  W.  J.  Britland  and  H.  E.  Potts i2mo,  *i  25 

Rollins,  W.     Notes  on  X-Light 8vo,  5  00 

Rollinson,  C.     Alphabets Oblong  i2mo,  1  00 

Rose,  J.     The  Pattern-makers'  Assistant 8vo,  2  50 

Key  to  Engines  and  Engine-running i2mo,  2  50 

Rose,  T.  K.     The  Precious  Metals.     (Westminster  Series.) .  .8vo,  *2  00 

Rosenhain,  W.  Glass  Manufacture.  (Westminster  Series.) .    8vo,  *2  00 

Ross,  W.  A.     Blowpipe  in  Chemistry  and  Metallurgy.  .  .i2mo,  *2  00 
Rossiter,  J.  T.     Steam  Engines.     (Westminster  Series.)  8vo  (In  Press.) 

Pumps  and  Pumping  Machinery.     (Westminster  Series.) 

8vo  (In  Press.) 

Roth.     Physical  Chemistry 8vo,  *2  00 

Rouillion,  L.     The  Economics  of  Manual  Training 8vo,  2  00 

Rowan,  F.  J.     Practical  Physics  of  the  Modern  Steam-boiler.8vo,  *3  00 


0 

go 

*3 

50 

*3 

50 

*3 

00 

i 

25 

i 

00 

D.  VAN  NOSTRAND  COMPANY'S  SHORT-TITLE  CATALOG    33 

Rowan,  F.  J.,  and  Idell,  F.  E.  Boiler  Incrustation  and  Corro- 
sion.     (Science  Series  No.  27.) i6mo, 

Roxburgh,  W.     General  Foundry  Practice 8vo, 

Ruhmer,  E.  Wireless  Telephony.  Trans,  by  J.  Erskine- 
Murray 8vo, 

Russell,  A.     Theory  of  Electric  Cables  and  Networks 8vo, 

Sabine,  R.  History  and  Progress  of  the  Electric  Telegraph.  i2mo, 

Saeltzer,  A.     Treatise  on  Acoustics i2mo, 

Salomons,  D.     Electric  Light  Installations.     i2mo. 

Vol.     I.     The  Management  of  Accumulators 2  50 

Electric  Light  Installations.     i2mo. 

Vol.    II.     Apparatus 

Vol.  III.     Applications 

Sanford,  P.  G.     Nitro-explosives 8vo, 

Saunders,  C.  H.     Handbook  of  Practical  Mechanics i6mo, 

leather, 

Saunnier,  C.     Watchmaker's  Handbook nmo, 

Sayers,  H.  M.     Brakes  for  Tram  Cars. 8vo, 

Scheele,  C.  W.     Chemical  Essays 8vo, 

Schellen,  H.     Magneto-electric  and  Dynamo -electric  Machines 

8vo, 

Scherer,  R.     Casein.     Trans,  by  C.  Salter 8vo, 

Schidrowitz,  P.     Rubber,  Its  Production  and  Uses 8vo, 

Schindler,  K.     Iron  and  Steel  Construction  Works i2mo, 

Schmall,  C.  N.     First  Course  in  Analytic  Geometry,  Plane  and 

Solid i2mo,  half  leather,     *i  75 

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2 

25 

I 

50 

*4 

00 

1 

00 

1 

25 

3 

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*2 

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5 

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*3 

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*3 

00 

*3 

00 

*3 

00 

2 

50 

7 

50 

:::5 

00 

Dre. 

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5 

00 

5 

00 

*2 

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*2 

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*2 

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50 

*2 

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*3 

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*3 

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*2 

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*3 

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*4 

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